ࡱ> eg!bcd_Vwfj h W 7 ]bjbjUU (7|7|h4llll8l ol3ZqV(~~~$- M  ~~g ~~>~Nq Ms`jl*\03n Chapter 10 1. The problem asks us to assume  EMBED Equation and w are constant. For consistency of units, we write  EMBED Equation  Thus, with EMBED Equation , the time of flight is  EMBED Equation . During that time, the angular displacement of a point on the ball s surface is  EMBED Equation  2. (a) The second hand of the smoothly running watch turns through 2p radians during  EMBED Equation.DSMT4 . Thus,  EMBED Equation.DSMT4  (b) The minute hand of the smoothly running watch turns through 2p radians during  EMBED Equation.DSMT4 . Thus,  EMBED Equation  (c) The hour hand of the smoothly running 12-hour watch turns through 2p radians during 43200 s. Thus,  EMBED Equation  3. The falling is the type of constant-acceleration motion you had in Chapter 2. The time it takes for the buttered toast to hit the floor is  EMBED Equation.DSMT4  (a) The smallest angle turned for the toast to land butter-side down is  EMBED Equation.DSMT4  This corresponds to an angular speed of  EMBED Equation.DSMT4  (b) The largest angle (less than 1 revolution) turned for the toast to land butter-side down is  EMBED Equation.DSMT4  This corresponds to an angular speed of  EMBED Equation.DSMT4  4. If we make the units explicit, the function is  EMBED Equation.DSMT4  but in some places we will proceed as indicated in the problem by letting these units be understood. (a) We evaluate the function q at t = 0 to obtain q0 = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 10-6:  EMBED Equation.DSMT4  which we evaluate at t = 0 to obtain w0 = 0. (c) For t = 4.0 s, the function found in the previous part is w4 = (8.0)(4.0) + (6.0)(4.0)2 = 128 rad/s. If we round this to two figures, we obtain w4 1.3 EMBED Equation.DSMT4 102 rad/s. (d) The angular acceleration as a function of time is given by Eq. 10-8:  EMBED Equation.DSMT4  which yields a2 = 8.0 + (12)(2.0) = 32 rad/s2 at t = 2.0 s. (e) The angular acceleration, given by the function obtained in the previous part, depends on time; it is not constant. 5. Applying Eq. 2-15 to the vertical axis (with +y downward) we obtain the free-fall time:  EMBED Equation  Thus, by Eq. 10-5, the magnitude of the average angular velocity is  EMBED Equation  6. If we make the units explicit, the function is  EMBED Equation  but generally we will proceed as shown in the problemletting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Equation 10-6 leads to  EMBED Equation  Evaluating this at t = 2 s yields w2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives w4 = 28 rad/s. (c) Consequently, Eq. 10-7 gives  EMBED Equation  (d) And Eq. 10-8 gives  EMBED Equation  Evaluating this at t = 2 s produces a2 = 6.0 rad/s2. (e) Evaluating the expression in part (d) at t = 4 s yields a4 = 18 rad/s2. We note that our answer for aavg does turn out to be the arithmetic average of a2 and a4 but point out that this will not always be the case. 7. (a) To avoid touching the spokes, the arrow must go through the wheel in not more than  EMBED Equation  The minimum speed of the arrow is then EMBED Equation  (b) No there is no dependence on radial position in the above computation. 8. (a) We integrate (with respect to time) the a = 6.0t4  4.0t2 expression, taking into account that the initial angular velocity is 2.0 rad/s. The result is w = 1.2 t5  1.33 t3 + 2.0. (b) Integrating again (and keeping in mind that qo = 1) we get q = 0.20t6  0.33 t4 + 2.0 t + 1.0 . 9. (a) With w = 0 and a =  4.2 rad/s2, Eq. 10-12 yields t =  wo/a = 3.00 s. (b) Eq. 10-4 gives q - qo = - wo2 / 2a = 18.9 rad. 10. We assume the sense of rotation is positive, which (since it starts from rest) means all quantities (angular displacements, accelerations, etc.) are positive-valued. (a) The angular acceleration satisfies Eq. 10-13:  EMBED Equation.DSMT4  (b) The average angular velocity is given by Eq. 10-5:  EMBED Equation  (c) Using Eq. 10-12, the instantaneous angular velocity at t = 5.0 s is  EMBED Equation  (d) According to Eq. 10-13, the angular displacement at t = 10 s is  EMBED Equation  Thus, the displacement between t = 5 s and t = 10 s is Dq = 100 rad  25 rad = 75 rad. 11. We assume the sense of initial rotation is positive. Then, with w0 = +120 rad/s and w = 0 (since it stops at time t), our angular acceleration (  deceleration  ) will be negative-valued: a =  4.0 rad/s2. (a) We apply Eq. 10-12 to obtain t.  EMBED Equation.DSMT4  (b) And Eq. 10-15 gives  EMBED Equation  Alternatively, Eq. 10-14 could be used if it is desired to only use the given information (as opposed to using the result from part (a)) in obtaining q. If using the result of part (a) is acceptable, then any angular equation in Table 10-1 (except Eq. 10-12) can be used to find q. 12. (a) We assume the sense of rotation is positive. Applying Eq. 10-12, we obtain  EMBED Equation.DSMT4  (b) And Eq. 10-15 gives  EMBED Equation =  EMBED Equation.DSMT4  rev. 13. The wheel has angular velocity w0 = +1.5 rad/s = +0.239 rev/s at t = 0, and has constant value of angular acceleration a < 0, which indicates our choice for positive sense of rotation. At t1 its angular displacement (relative to its orientation at t = 0) is q1 = +20 rev, and at t2 its angular displacement is q2 = +40 rev and its angular velocity is  EMBED Equation.DSMT4 . (a) We obtain t2 using Eq. 10-15:  EMBED Equation  which we round off to  EMBED Equation.DSMT4 . (b) Any equation in Table 10-1 involving a can be used to find the angular acceleration; we select Eq. 10-16.  EMBED Equation.DSMT4  which we convert to a =  4.5 10 3 rad/s2. (c) Using  EMBED Equation  (Eq. 10-13) and the quadratic formula, we have  EMBED Equation  which yields two positive roots: 98 s and 572 s. Since the question makes sense only if t1 < t2 we conclude the correct result is t1 = 98 s. 14. The wheel starts turning from rest (w0 = 0) at t = 0, and accelerates uniformly at a > 0, which makes our choice for positive sense of rotation. At t1 its angular velocity is w1 = +10 rev/s, and at t2 its angular velocity is w2 = +15 rev/s. Between t1 and t2 it turns through Dq = 60 rev, where t2  t1 = Dt. (a) We find a using Eq. 10-14:  EMBED Equation.DSMT4  which we round off to 1.0 rev/s2. (b) We find Dt using Eq. 10-15:  EMBED Equation  (c) We obtain t1 using Eq. 10-12:  EMBED Equation.DSMT4  (d) Any equation in Table 10-1 involving q can be used to find q1 (the angular displacement during 0 t t1); we select Eq. 10-14.  EMBED Equation.DSMT4  15. We have a wheel rotating with constant angular acceleration. We can apply the equations given in Table 10-1 to analyze the motion. Since the wheel starts from rest, its angular displacement as a function of time is given by  EMBED Equation.DSMT4 . We take  EMBED Equation.DSMT4  to be the start time of the interval so that  EMBED Equation.DSMT4 . The corresponding angular displacements at these times are  EMBED Equation  Given  EMBED Equation.DSMT4 , we can solve for  EMBED Equation.DSMT4 , which tells us how long the wheel has been in motion up to the beginning of the 4.0 s-interval. The above expressions can be combined to give  EMBED Equation  With  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 , and  EMBED Equation.DSMT4 , we obtain  EMBED Equation , which can be further solved to give  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 . So, the wheel started from rest 8.0 s before the start of the described 4.0 s interval. Note: We can readily verify the results by calculating  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4  explicitly:  EMBED Equation  Indeed the difference is  EMBED Equation.DSMT4 . 16. (a) Eq. 10-13 gives q - qo = wo t +  EQ \f(1,2) at2 = 0 +  EQ \f(1,2) (1.5 rad/s) t12 where q - qo = (2 rev)(2p rad/rev). Therefore, t1 = 4.09 s. (b) We can find the time to go through a full 4 rev (using the same equation to solve for a new time t2) and then subtract the result of part (a) for t1 in order to find this answer. (4 rev)(2p rad/rev) = 0 +  EQ \f(1,2) (1.5 rad/s) t22 ( t2 = 5.789 s. Thus, the answer is 5.789 s 4.093 s ( 1.70 s. 17. The problem has (implicitly) specified the positive sense of rotation. The angular acceleration of magnitude 0.25 rad/s2 in the negative direction is assumed to be constant over a large time interval, including negative values (for t). (a) We specify qmax with the condition w = 0 (this is when the wheel reverses from positive rotation to rotation in the negative direction). We obtain qmax using Eq. 10-14:  EMBED Equation  (b) We find values for t1 when the angular displacement (relative to its orientation at t = 0) is q1 = 22 rad (or 22.09 rad if we wish to keep track of accurate values in all intermediate steps and only round off on the final answers). Using Eq. 10-13 and the quadratic formula, we have  EMBED Equation.DSMT4  which yields the two roots 5.5 s and 32 s. Thus, the first time the reference line will be at q1 = 22 rad is t = 5.5 s. (c) The second time the reference line will be at q1 = 22 rad is t = 32 s. (d) We find values for t2 when the angular displacement (relative to its orientation at t = 0) is q2 =  10.5 rad. Using Eq. 10-13 and the quadratic formula, we have  EMBED Equation.DSMT4  which yields the two roots  2.1 s and 40 s. Thus, at t =  2.1 s the reference line will be at q2 =  10.5 rad. (e) At t = 40 s the reference line will be at q2 =  10.5 rad. (f) With radians and seconds understood, the graph of q versus t is shown below (with the points found in the previous parts indicated as small dots).  18. First, we convert the angular velocity: w = (2000 rev/min)(2p /60) = 209 rad/s. Also, we convert the plane s speed to SI units: (480)(1000/3600) = 133 m/s. We use Eq. 10-18 in part (a) and (implicitly) Eq. 4-39 in part (b). (a) The speed of the tip as seen by the pilot is EMBED Equation , which (since the radius is given to only two significant figures) we write as  EMBED Equation.DSMT4 . (b) The planes velocity  EMBED Equation  and the velocity of the tip  EMBED Equation  (found in the planes frame of reference), in any of the tips positions, must be perpendicular to each other. Thus, the speed as seen by an observer on the ground is  EMBED Equation  19. (a) Converting from hours to seconds, we find the angular velocity (assuming it is positive) from Eq. 10-18:  EMBED Equation  (b) The radial (or centripetal) acceleration is computed according to Eq. 10-23:  EMBED Equation  (c) Assuming the angular velocity is constant, then the angular acceleration and the tangential acceleration vanish, since  EMBED Equation  20. The function  EMBED Equation  where x = 0.40 rad and b = 2 s 1 is describing the angular coordinate of a line (which is marked in such a way that all points on it have the same value of angle at a given time) on the object. Taking derivatives with respect to time leads to  EMBED Equation  and  EMBED Equation  (a) Using Eq. 10-22, we have  EMBED Equation  (b) Using Eq. 10-23, we get  EMBED Equation  21. We assume the given rate of 1.2 10 3 m/y is the linear speed of the top; it is also possible to interpret it as just the horizontal component of the linear speed but the difference between these interpretations is arguably negligible. Thus, Eq. 10-18 leads to  EMBED Equation  which we convert (since there are about 3.16 107 s in a year) to w = 6.9 10 13 rad/s. 22. (a) Using Eq. 10-6, the angular velocity at t = 5.0s is  EMBED Equation  (b) Equation 10-18 gives the linear speed at t = 5.0s:  EMBED Equation.DSMT4  (c) The angular acceleration is, from Eq. 10-8,  EMBED Equation  Then, the tangential acceleration at t = 5.0s is, using Eq. 10-22,  EMBED Equation  (d) The radial (centripetal) acceleration is given by Eq. 10-23:  EMBED Equation  23. The linear speed of the flywheel is related to its angular speed by  EMBED Equation.DSMT4 , where r is the radius of the wheel. As the wheel is accelerated, its angular speed at a later time is  EMBED Equation.DSMT4 . (a) The angular speed of the wheel, expressed in rad/s, is  EMBED Equation  (b) With r = (1.20 m)/2 = 0.60 m, using Eq. 10-18, we find the linear speed to be  EMBED Equation.DSMT4  (c) With t = 1 min, w = 1000 rev/min and w0 = 200 rev/min, Eq. 10-12 gives the required acceleration:  EMBED Equation  (d) With the same values used in part (c), Eq. 10-15 becomes  EMBED Equation  Note: An alternative way to solve for (d) is to use Eq. 10-13:  EMBED Equation  24. Converting 33 EQ \f(1,3)  rev/min to radians-per-second, we get w = 3.49 rad/s. Combining  EMBED Equation.DSMT4 (Eq. 10-18) with Dt = d/v where Dt is the time between bumps (a distance d apart), we arrive at the rate of striking bumps:  EMBED Equation.DSMT4 . 25. The linear speed of a point on Earth s surface depends on its distance from the axis of rotation. To solve for the linear speed, we use v = w r, where r is the radius of its orbit. A point on Earth at a latitude of 40 moves along a circular path of radius r = R cos 40, where R is the radius of Earth (6.4 106 m). On the other hand, r = R at the equator. (a) Earth makes one rotation per day and 1 d is (24 h) (3600 s/h) = 8.64 104 s, so the angular speed of Earth is  EMBED Equation  (b) At latitude of 40, the linear speed is  EMBED Equation  (c) At the equator (and all other points on Earth) the value of w is the same (7.3 10 5 rad/s). (d) The latitude at the equator is 0 and the speed is  EMBED Equation  Note: The linear speed at the poles is zero since  EMBED Equation.DSMT4 . 26. (a) The angular acceleration is  EMBED Equation  (b) Using Eq. 10-13 with t = (2.2) (60) = 132 min, the number of revolutions is  EMBED Equation.DSMT4  (c) With r = 500 mm, the tangential acceleration is  EMBED Equation  which yields at = 0.99 mm/s2. (d) The angular speed of the flywheel is  EMBED Equation  With r = 0.50 m, the radial (or centripetal) acceleration is given by Eq. 10-23:  EMBED Equation  which is much bigger than at. Consequently, the magnitude of the acceleration is  EMBED Equation  27. (a) The angular speed in rad/s is  EMBED Equation  Consequently, the radial (centripetal) acceleration is (using Eq. 10-23)  EMBED Equation  (b) Using Ch. 6 methods, we have ma = fs fs,max = ms mg, which is used to obtain the (minimum allowable) coefficient of friction:  EMBED Equation  (c) The radial acceleration of the object is ar = w2r, while the tangential acceleration is at = ar. Thus,  EMBED Equation  If the object is not to slip at any time, we require  EMBED Equation  Thus, since a = w/t (from Eq. 10-12), we find  EMBED Equation  28. Since the belt does not slip, a point on the rim of wheel C has the same tangential acceleration as a point on the rim of wheel A. This means that aArA = aCrC, where aA is the angular acceleration of wheel A and aC is the angular acceleration of wheel C. Thus,  EMBED Equation  With the angular speed of wheel C given by  EMBED Equation.DSMT4 , the time for it to reach an angular speed of w = 100 rev/min = 10.5 rad/s starting from rest is  EMBED Equation  29. (a) In the time light takes to go from the wheel to the mirror and back again, the wheel turns through an angle of q = 2p/500 = 1.26 10 2 rad. That time is  EMBED Equation  so the angular velocity of the wheel is  EMBED Equation  (b) If r is the radius of the wheel, the linear speed of a point on its rim is  EMBED Equation.DSMT4  30. (a) The tangential acceleration, using Eq. 10-22, is  EMBED Equation  (b) In rad/s, the angular velocity is w = (2760)(2p/60) = 289 rad/s, so  EMBED Equation  (c) The angular displacement is, using Eq. 10-14,  EMBED Equation  Then, using Eq. 10-1, the distance traveled is  EMBED Equation  31. (a) The upper limit for centripetal acceleration (same as the radial acceleration  see Eq. 10-23) places an upper limit of the rate of spin (the angular velocity w) by considering a point at the rim (r = 0.25 m). Thus, wmax =  EQ \r(a/r)  = 40 rad/s. Now we apply Eq. 10-15 to first half of the motion (where wo = 0): q - qo =  EQ \f(1,2) (wo + w)t  EMBED Equation.DSMT4  400 rad =  EQ \f(1,2) (0 + 40 rad/s)t which leads to t = 20 s. The second half of the motion takes the same amount of time (the process is essentially the reverse of the first); the total time is therefore 40 s. (b) Considering the first half of the motion again, Eq. 10-11 leads to w = wo + a t ( a =  EQ \f(40 rad/s,20 s)  = 2.0 rad/s2 . 32. (a) A complete revolution is an angular displacement of Dq = 2p rad, so the angular velocity in rad/s is given by w = Dq/T = 2p/T. The angular acceleration is given by  EMBED Equation  For the pulsar described in the problem, we have  EMBED Equation  Therefore,  EMBED Equation  The negative sign indicates that the angular acceleration is opposite the angular velocity and the pulsar is slowing down. (b) We solve w = w0 + at for the time t when w = 0:  EMBED Equation.DSMT4  (c) The pulsar was born 1992 1054 = 938 years ago. This is equivalent to (938 y)(3.16 107 s/y) = 2.96 1010 s. Its angular velocity at that time was  EMBED Equation.DSMT4  Its period was  EMBED Equation  33. The kinetic energy (in J) is given by  EMBED Equation  where I is the rotational inertia (in  EMBED Equation ) and w is the angular velocity (in rad/s). We have  EMBED Equation  Consequently, the rotational inertia is  EMBED Equation  34. (a) Equation 10-12 implies that the angular acceleration a should be the slope of the w vs t graph. Thus, a = 9/6 = 1.5 rad/s2. (b) By Eq. 10-34, K is proportional to w2. Since the angular velocity at t = 0 is  2 rad/s (and this value squared is 4) and the angular velocity at t = 4 s is 4 rad/s (and this value squared is 16), then the ratio of the corresponding kinetic energies must be  EQ \f(Ko, K4)  =  EQ \f(4,16)  ( Ko = K4/4 = 0.40 J . 35. Since the rotational inertia of a cylinder is  EMBED Equation  (Table 10-2(c)), its rotational kinetic energy is  EMBED Equation.DSMT4  (a) For the smaller cylinder, we have  EMBED Equation  (b) For the larger cylinder, we obtain  EMBED Equation  36. The parallel axis theorem (Eq. 10-36) shows that I increases with h. The phrase out to the edge of the disk (in the problem statement) implies that the maximum h in the graph is, in fact, the radius R of the disk. Thus, R = 0.20 m. Now we can examine, say, the h = 0 datum and use the formula for Icom (see Table 10-2(c)) for a solid disk, or (which might be a little better, since this is independent of whether it is really a solid disk) we can the difference between the h = 0 datum and the h = hmax =R datum and relate that difference to the parallel axis theorem (thus the difference is M(hmax)2 = 0.10  EMBED Equation.DSMT4 ). In either case, we arrive at M = 2.5 kg. 37. We use the parallel axis theorem: I = Icom + Mh2, where Icom is the rotational inertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m 0.20 m = 0.30 m. We find  EMBED Equation  Consequently, the parallel axis theorem yields  EMBED Equation  38. (a) Equation 10-33 gives Itotal = md2 + m(2d)2 + m(3d)2 = 14 md2. If the innermost one is removed then we would only obtain m(2d)2 + m(3d)2 = 13 md2. The percentage difference between these is (13 14)/14 = 0.0714 ( 7.1%. (b) If, instead, the outermost particle is removed, we would have md2 + m(2d)2 = 5 md2. The percentage difference in this case is 0.643 ( 64%. 39. (a) Using Table 10-2(c) and Eq. 10-34, the rotational kinetic energy is  EMBED Equation  (b) We solve P = K/t (where P is the average power) for the operating time t.  EMBED Equation  which we rewrite as t 1.0 102 min. 40. (a) Consider three of the disks (starting with the one at point O): (OO . The first one (the one at point O, shown here with the plus sign inside) has rotational inertial (see item (c) in Table 10-2) I =  EQ \f(1,2) mR2. The next one (using the parallel-axis theorem) has I =  EQ \f(1,2) mR2 + mh2 where h = 2R. The third one has I =  EQ \f(1,2) mR2 + m(4R)2. If we had considered five of the disks OO(OO with the one at O in the middle, then the total rotational inertia is I = 5( EQ \f(1,2) mR2) + 2(m(2R)2 + m(4R)2). The pattern is now clear and we can write down the total I for the collection of fifteen disks: I = 15( EQ \f(1,2) mR2) + 2(m(2R)2 + m(4R)2 + m(6R)2+ + m(14R)2) =  EQ \f(2255,2) mR2. The generalization to N disks (where N is assumed to be an odd number) is I =  EQ \f(1,6)(2N2 + 1)NmR2. In terms of the total mass (m = M/15) and the total length (R = L/30), we obtain I = 0.083519ML2 ( (0.08352)(0.1000 kg)(1.0000 m)2 = 8.352 10-3 kg' m2. (b) Comparing to the formula (e) in Table 10-2 (which gives roughly I =0.08333 ML2), we find our answer to part (a) is 0.22% lower. 41. The particles are treated  point-like in the sense that Eq. 10-33 yields their rotational inertia, and the rotational inertia for the rods is figured using Table 10-2(e) and the parallel-axis theorem (Eq. 10-36). (a) With subscript 1 standing for the rod nearest the axis and 4 for the particle farthest from it, we have  EMBED Equation  (b) Using Eq. 10-34, we have  EMBED Equation  42. (a) We apply Eq. 10-33:  EMBED Equation  (b) For rotation about the y axis we obtain  EMBED Equation  (c) And about the z axis, we find (using the fact that the distance from the z axis is  EMBED Equation )  EMBED Equation  (d) Clearly, the answer to part (c) is A + B. 43. Since the rotation axis does not pass through the center of the block, we use the parallel-axis theorem to calculate the rotational inertia. According to Table 10-2(i), the rotational inertia of a uniform slab about an axis through the center and perpendicular to the large faces is given by  EMBED Equation  A parallel axis through the corner is a distance  EMBED Equation  from the center. Therefore,  EMBED Equation  With  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , we have  EMBED Equation  44. (a) We show the figure with its axis of rotation (the thin horizontal line).  We note that each mass is r = 1.0 m from the axis. Therefore, using Eq. 10-26, we obtain  EMBED Equation.DSMT4  (b) In this case, the two masses nearest the axis are r = 1.0 m away from it, but the two furthest from the axis are  EMBED Equation  from it. Here, then, Eq. 10-33 leads to  EMBED Equation  (c) Now, two masses are on the axis (with r = 0) and the other two are a distance  EMBED Equation  away. Now we obtain  EMBED Equation.DSMT4  45. We take a torque that tends to cause a counterclockwise rotation from rest to be positive and a torque tending to cause a clockwise rotation to be negative. Thus, a positive torque of magnitude r1 F1 sin q1 is associated with  EMBED Equation  and a negative torque of magnitude r2F2 sin q2 is associated with  EMBED Equation . The net torque is consequently  EMBED Equation  Substituting the given values, we obtain  EMBED Equation.DSMT4  46. The net torque is  EMBED Equation.DSMT4  47. Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot point is associated with the force of the rod since that force is along the line from the pivot point to the ball.  As can be seen from the diagram, the component of the force of gravity that is perpendicular to the rod is mg sin q. If  EMBED Equation  is the length of the rod, then the torque associated with this force has magnitude  EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4 . For the position shown, the torque is counterclockwise. 48. We compute the torques using t = rF sin f. (a) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 . (b) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 . (c) For  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 . 49. (a) We use the kinematic equation  EMBED Equation , where w0 is the initial angular velocity, w is the final angular velocity, a is the angular acceleration, and t is the time. This gives  EMBED Equation  (b) If I is the rotational inertia of the diver, then the magnitude of the torque acting on her is  EMBED Equation  50. The rotational inertia is found from Eq. 10-45.  EMBED Equation  51. (a) We use constant acceleration kinematics. If down is taken to be positive and a is the acceleration of the heavier block m2, then its coordinate is given by  EMBED Equation , so  EMBED Equation  Block 1 has an acceleration of 6.00 10 2 m/s2 upward. (b) Newton s second law for block 2 is  EMBED Equation , where m2 is its mass and T2 is the tension force on the block. Thus,  EMBED Equation  (c) Newton s second law for block 1 is  EMBED Equation  where T1 is the tension force on the block. Thus,  EMBED Equation  (d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of the pulley must be the same as the acceleration of the blocks, so  EMBED Equation  (e) The net torque acting on the pulley is  EMBED Equation.DSMT4 . Equating this to Ia we solve for the rotational inertia:  EMBED Equation.DSMT4  52. According to the sign conventions used in the book, the magnitude of the net torque exerted on the cylinder of mass m and radius R is  EMBED Equation  (a) The resulting angular acceleration of the cylinder (with  EMBED Equation  according to Table 10-2(c)) is  EMBED Equation.DSMT4 . (b) The direction is counterclockwise (which is the positive sense of rotation). 53. Combining Eq. 10-45 (tnet = I a) with Eq. 10-38 gives RF2  RF1 = Ia , where  EMBED Equation.DSMT4  by Eq. 10-12 (with wo = 0). Using item (c) in Table 10-2 and solving for F2 we find F2 =  EMBED Equation.DSMT4  + F1 =  EQ \f((0.02)(0.02)(250),2(1.25))  + 0.1 = 0.140 N. 54. (a) In this case, the force is mg = (70 kg)(9.8 m/s2), and the lever arm (the perpendicular distance from point O to the line of action of the force) is 0.28 m. Thus, the torque (in absolute value) is (70 kg)(9.8 m/s2)(0.28 m). Since the moment-of-inertia is I = 65  EMBED Equation.DSMT4 , then Eq. 10-45 gives |a| = 2.955 ( 3.0 rad/s2. (b) Now we have another contribution (1.4 m ( 300 N) to the net torque, so |tnet| = (70 kg)(9.8 m/s2)(0.28 m) + (1.4 m)(300 N) = (65  EMBED Equation.DSMT4 ) |a| which leads to |a| = 9.4 rad/s2. 55. Combining Eq. 10-34 and Eq. 10-45, we have RF = Ia, where a is given by w/t (according to Eq. 10-12, since wo = 0 in this case). We also use the fact that I = Iplate + Idisk where Idisk =  EQ \f(1,2) MR2 (item (c) in Table 10-2). Therefore, Iplate =  EQ \f(RFt, w)    EQ \f(1,2) MR2 = 2.51 ( 10-4  EMBED Equation.DSMT4 . 56. With counterclockwise positive, the angular acceleration a for both masses satisfies  EMBED Equation.DSMT4  by combining Eq. 10-45 with Eq. 10-39 and Eq. 10-33. Therefore, using SI units,  EMBED Equation  where the negative sign indicates the system starts turning in the clockwise sense. The magnitude of the acceleration vector involves no radial component (yet) since it is evaluated at t = 0 when the instantaneous velocity is zero. Thus, for the two masses, we apply Eq. 10-22: (a)  EMBED Equation.DSMT4  (b)  EMBED Equation.DSMT4  57. Since the force acts tangentially at r = 0.10 m, the angular acceleration (presumed positive) is  EMBED Equation  in SI units (rad/s2). (a) At t = 3 s, the above expression becomes a = 4.2 102 rad/s2. (b) We integrate the above expression, noting that wo = 0, to obtain the angular speed at t = 3 s:  EMBED Equation.DSMT4  58. (a) The speed of v of the mass m after it has descended d = 50 cm is given by v2 = 2ad (Eq. 2-16). Thus, using g = 980 cm/s2, we have EMBED Equation (b) The answer is still 1.4 102 cm/s = 1.4 m/s, since it is independent of R. 59. With w = (1800)(2p/60) = 188.5 rad/s, we apply Eq. 10-55:  EMBED Equation.DSMT4 . 60. (a) We apply Eq. 10-34:  EMBED Equation  (b) Simple conservation of mechanical energy leads to K = mgh. Consequently, the center of mass rises by  EMBED Equation  61. The initial angular speed is w = (280 rev/min)(2p/60) = 29.3 rad/s. (a) Since the rotational inertia is (Table 10-2(a))  EMBED Equation , the work done is  EMBED Equation  . (b) The average power (in absolute value) is therefore  EMBED Equation  62. (a) Eq. 10-33 gives Itotal = md2 + m(2d)2 + m(3d)2 = 14 md2, where d = 0.020 m and m = 0.010 kg. The work done is W = DK =  EQ \f(1,2) Iwf 2   EQ \f(1,2) Iwi2, where wf = 20 rad/s and wi = 0. This gives W = 11.2 mJ. (b) Now, wf = 40 rad/s and wi = 20 rad/s, and we get W = 33.6 mJ. (c) In this case, wf = 60 rad/s and wi = 40 rad/s. This gives W = 56.0 mJ. (d) Equation 10-34 indicates that the slope should be  EQ \f(1,2) I. Therefore, it should be 7md2 = 2.80 ( 10-5 J.s2/ rad2. 63. We use  EMBED Equation  to denote the length of the stick. Since its center of mass is  EMBED Equation  from either end, its initial potential energy is  EMBED Equation  where m is its mass. Its initial kinetic energy is zero. Its final potential energy is zero, and its final kinetic energy is  EMBED Equation  where I is its rotational inertia about an axis passing through one end of the stick and w is the angular velocity just before it hits the floor. Conservation of energy yields  EMBED Equation  The free end of the stick is a distance  EMBED Equation  from the rotation axis, so its speed as it hits the floor is (from Eq. 10-18)  EMBED Equation  Using Table 10-2 and the parallel-axis theorem, the rotational inertial is  EMBED Equation , so  EMBED Equation  64. (a) We use the parallel-axis theorem to find the rotational inertia:  EMBED Equation.DSMT4  (b) Conservation of energy requires that  EMBED Equation , where w is the angular speed of the cylinder as it passes through the lowest position. Therefore,  EMBED Equation  65. (a) We use conservation of mechanical energy to find an expression for w2 as a function of the angle q that the chimney makes with the vertical. The potential energy of the chimney is given by U = Mgh, where M is its mass and h is the altitude of its center of mass above the ground. When the chimney makes the angle q with the vertical, h = (H/2) cos q. Initially the potential energy is Ui = Mg(H/2) and the kinetic energy is zero. The kinetic energy is  EMBED Equation  when the chimney makes the angle q with the vertical, where I is its rotational inertia about its bottom edge. Conservation of energy then leads to  EMBED Equation  The rotational inertia of the chimney about its base is I = MH2/3 (found using Table 10-2(e) with the parallel axis theorem). Thus  EMBED Equation  (b) The radial component of the acceleration of the chimney top is given by ar = Hw2, so ar = 3g (1  cos q) = 3 (9.80 m/s2)(1 cos 35.0 EMBED Equation.DSMT4 ) = 5.32 m/s2 . (c) The tangential component of the acceleration of the chimney top is given by at = Ha, where a is the angular acceleration. We are unable to use Table 10-1 since the acceleration is not uniform. Hence, we differentiate w2 = (3g/H)(1  cos q) with respect to time, replacing dw / dt with a, and dq / dt with w, and obtain  EMBED Equation  Consequently,  EMBED Equation  (d) The angle q at which at = g is the solution to  EMBED Equation  Thus, sin q = 2/3 and we obtain q = 41.8. 66. From Table 10-2, the rotational inertia of the spherical shell is 2MR2/3, so the kinetic energy (after the object has descended distance h) is  EMBED Equation  Since it started from rest, then this energy must be equal (in the absence of friction) to the potential energy mgh with which the system started. We substitute v/r for the pulleys angular speed and v/R for that of the sphere and solve for v.  EMBED Equation  67. Using the parallel axis theorem and items (e) and (h) in Table 10-2, the rotational inertia is I =  EQ \f(1,12) mL2 + m(L/2)2 +  EQ \f(1,2) mR2 + m(R + L)2 = 10.83mR2 , where L = 2R has been used. If we take the base of the rod to be at the coordinate origin (x = 0, y = 0) then the center of mass is at y =  EQ \f(mL/2 + m(L + R),m + m)  = 2R . Comparing the position shown in the textbook figure to its upside down (inverted) position shows that the change in center of mass position (in absolute value) is |Dy| = 4R. The corresponding loss in gravitational potential energy is converted into kinetic energy. Thus, K = (2m)g(4R) ( w = 9.82 rad/s where Eq. 10-34 has been used. 68. We choose directions such that the initial angular velocity is w0 =  317 rad/s and the values for a, t, and F are positive. (a) Combining Eq. 10-12 with Eq. 10-45 and Table 10-2(f) (and using the fact that w = 0) we arrive at the expression  EMBED Equation  With t = 15.5 s, R = 0.226 m, and M = 1.65 kg, we obtain t = 0.689 N m. (b) From Eq. 10-40, we find F = t /R = 3.05 N. (c) Using again the expression found in part (a), but this time with R = 0.854 m, we get  EMBED Equation.DSMT4 . (d) Now, F = t / R = 11.5 N. 69. The volume of each disk is pr2h where we are using h to denote the thickness (which equals 0.00500 m). If we use R (which equals 0.0400 m) for the radius of the larger disk and r (which equals 0.0200 m) for the radius of the smaller one, then the mass of each is m = rpr2h and M = rpR2h where r = 1400 kg/m3 is the given density. We now use the parallel axis theorem as well as item (c) in Table 10-2 to obtain the rotation inertia of the two-disk assembly: I =  EQ \f(1,2) MR2 +  EQ \f(1,2) mr2 + m(r + R)2 = rph[  EQ \f(1,2) R4 +  EQ \f(1,2) r4 + r2(r + R)2 ] = 6.16 ( 10-5  EMBED Equation.DSMT4 . 70. The wheel starts turning from rest (w0 = 0) at t = 0, and accelerates uniformly at  EMBED Equation.DSMT4 . Between t1 and t2 the wheel turns through Dq = 90.0 rad, where t2  t1 = Dt = 3.00 s. We solve (b) first. (b) We use Eq. 10-13 (with a slight change in notation) to describe the motion for t1 t t2:  EMBED Equation  which we plug into Eq. 10-12, set up to describe the motion during 0 t t1:  EMBED Equation.DSMT4  yielding t1 = 13.5 s. (a) Plugging into our expression for w1 (in previous part) we obtain  EMBED Equation  71. We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set a2 = a1 = Ra (for simplicity, we denote this as a). Thus, we choose rightward positive for m2 = M (the block on the table), downward positive for m1 = M (the block at the end of the string) and (somewhat unconventionally) clockwise for positive sense of disk rotation. This means that we interpret q given in the problem as a positive-valued quantity. Applying Newton s second law to m1, m2 and (in the form of Eq. 10-45) to M, respectively, we arrive at the following three equations (where we allow for the possibility of friction f2 acting on m2).  EMBED Equation  (a) From Eq. 10-13 (with w0 = 0) we find  EMBED Equation  (b) From the fact that a = Ra (noted above), we obtain  EMBED Equation.DSMT4  (c) From the first of the above equations, we find  EMBED Equation  (d) From the last of the above equations, we obtain the second tension:  EMBED Equation  72. (a) Constant angular acceleration kinematics can be used to compute the angular acceleration a. If w0 is the initial angular velocity and t is the time to come to rest, then  EMBED Equation.DSMT4 , which gives  EMBED Equation  . (b) We use t = Ia, where t is the torque and I is the rotational inertia. The contribution of the rod to I is  EMBED Equation  (Table 10-2(e)), where M is its mass and  EMBED Equation  is its length. The contribution of each ball is  EMBED Equation  where m is the mass of a ball. The total rotational inertia is  EMBED Equation  which yields I = 1.53 kg EMBED Equation m2. The torque, therefore, is  EMBED Equation  (c) Since the system comes to rest the mechanical energy that is converted to thermal energy is simply the initial kinetic energy  EMBED Equation  (d) We apply Eq. 10-13:  EMBED Equation  which yields 3920 rad or (dividing by 2p) 624 rev for the value of angular displacement q. (e) Only the mechanical energy that is converted to thermal energy can still be computed without additional information. It is 4.59 104 J no matter how t varies with time, as long as the system comes to rest. 73. The Hint given in the problem would make the computation in part (a) very straightforward (without doing the integration as we show here), but we present this further level of detail in case that hint is not obvious or simply in case one wishes to see how the calculus supports our intuition. (a) The (centripetal) force exerted on an infinitesimal portion of the blade with mass dm located a distance r from the rotational axis is (Newtons second law) dF = (dm)w2r, where dm can be written as (M/L)dr and the angular speed is  EMBED Equation.DSMT4  EMBED Equation.DSMT4 . Thus for the entire blade of mass M and length L the total force is given by  EMBED Equation  (b) About its center of mass, the blade has  EMBED Equation  according to Table 10-2(e), and using the parallel-axis theorem to move the axis of rotation to its end-point, we find the rotational inertia becomes  EMBED Equation  Using Eq. 10-45, the torque (assumed constant) is  EMBED Equation  (c) Using Eq. 10-52, the work done is  EMBED Equation  74. The angular displacements of disks A and B can be written as:  EMBED Equation.DSMT4  (a) The time when  EMBED Equation.DSMT4  is given by  EMBED Equation.DSMT4  (b) The difference in the angular displacement is  EMBED Equation.DSMT4  For their reference lines to align momentarily, we only require  EMBED Equation.DSMT4 , where N is an integer. The quadratic equation can be readily solve to yield  EMBED Equation.DSMT4  The solution  EMBED Equation.DSMT4 (taking the positive root) coincides with the result obtained in (a), while  EMBED Equation.DSMT4 (taking the negative root) is the moment when both disks begin to rotate. In fact, two solutions exist for N = 0, 1, 2, and 3. 75. The magnitude of torque is the product of the force magnitude and the distance from the pivot to the line of action of the force. In our case, it is the gravitational force that passes through the walkers center of mass. Thus,  EMBED Equation.DSMT4  (a) Without the pole, with EMBED Equation.DSMT4 , the angular acceleration is  EMBED Equation.DSMT4  (b) When the walker carries a pole, the torque due to the gravitational force through the poles center of mass opposes the torque due to the gravitational force that passes through the walkers center of mass. Therefore,  EMBED Equation.DSMT4 , and the resulting angular acceleration is  EMBED Equation.DSMT4  76. The motion consists of two stages. The first, the interval 0 t 20 s, consists of constant angular acceleration given by  EMBED Equation  The second stage, 20 < t 40 s, consists of constant angular velocity  EMBED Equation  Analyzing the first stage, we find  EMBED Equation.DSMT4  Analyzing the second stage, we obtain  EMBED Equation  77. We assume the sense of initial rotation is positive. Then, with w0 > 0 and w = 0 (since it stops at time t), our angular acceleration is negative-valued. (a) The angular acceleration is constant, so we can apply Eq. 10-12 (w = w0 + at). To obtain the requested units, we have t = 30/60 = 0.50 min. Thus,  EMBED Equation  (b) We use Eq. 10-13:  EMBED Equation  78. We use conservation of mechanical energy. The center of mass is at the midpoint of the cross bar of the H and it drops by L/2, where L is the length of any one of the rods. The gravitational potential energy decreases by MgL/2, where M is the mass of the body. The initial kinetic energy is zero and the final kinetic energy may be written  EMBED Equation , where I is the rotational inertia of the body and w is its angular velocity when it is vertical. Thus,  EMBED Equation  Since the rods are thin the one along the axis of rotation does not contribute to the rotational inertia. All points on the other leg are the same distance from the axis of rotation, so that leg contributes (M/3)L2, where M/3 is its mass. The cross bar is a rod that rotates around one end, so its contribution is (M/3)L2/3 = ML2/9. The total rotational inertia is I = (ML2/3) + (ML2/9) = 4ML2/9. Consequently, the angular velocity is  EMBED Equation  79. (a) According to Table 10-2, the rotational inertia formulas for the cylinder (radius R) and the hoop (radius r) are given by  EMBED Equation  Since the two bodies have the same mass, then they will have the same rotational inertia if  EMBED Equation  !  EMBED Equation . (b) We require the rotational inertia to be written as  EMBED Equation , where M is the mass of the given body and k is the radius of the  equivalent hoop. It follows directly that  EMBED Equation . 80. (a) Using Eq. 10-15, we have 60.0 rad =  EQ \f(1,2) (w1 + w2)(6.00 s) . With w2 = 15.0 rad/s, then w1 = 5.00 rad/s. (b) Eq. 10-12 gives a = (15.0 rad/s  5.0 rad/s)/(6.00 s) = 1.67 rad/s2. (c) Interpreting w now as w1 and q as q1 = 10.0 rad (and wo = 0) Eq. 10-14 leads to qo =   EMBED Equation.DSMT4  + q1 = 2.50 rad . 81. The center of mass is initially at height  EMBED Equation  when the system is released (where L = 2.0 m). The corresponding potential energy Mgh (where M = 1.5 kg) becomes rotational kinetic energy  EMBED Equation  as it passes the horizontal position (where I is the rotational inertia about the pin). Using Table 10-2 (e) and the parallel axis theorem, we find  EMBED Equation  Therefore,  EMBED Equation.DSMT4  82. The rotational inertia of the passengers is (to a good approximation) given by Eq. 10-53:  EMBED Equation  where N is the number of people and m is the (estimated) mass per person. We apply Eq. 10-52:  EMBED Equation.DSMT4  where R = 38 m and N = 36 60 = 2160 persons. The rotation rate is constant so that w = q/t which leads to w = 2p/120 = 0.052 rad/s. The mass (in kg) of the average person is probably in the range 50 m 100, so the work should be in the range  EMBED Equation  83. We choose positive coordinate directions (different choices for each item) so that each is accelerating positively, which will allow us to set  EMBED Equation  (for simplicity, we denote this as a). Thus, we choose upward positive for m1, downward positive for m2, and (somewhat unconventionally) clockwise for positive sense of disk rotation. Applying Newtons second law to m1m2 and (in the form of Eq. 10-45) to M, respectively, we arrive at the following three equations.  EMBED Equation  (a) The rotational inertia of the disk is  EMBED Equation  (Table 10-2(c)), so we divide the third equation (above) by R, add them all, and use the earlier equality among accelerations to obtain:  EMBED Equation  which yields  EMBED Equation.DSMT4  (b) Plugging back in to the first equation, we find  EMBED Equation  where it is important in this step to have the mass in SI units: m1 = 0.40 kg. (c) Similarly, with m2 = 0.60 kg, we find  EMBED Equation  84. (a) The longitudinal separation between Helsinki and the explosion site is  EMBED Equation  The spin of the Earth is constant at  EMBED Equation  so that an angular displacement of  EMBED Equation  corresponds to a time interval of  EMBED Equation  (b) Now  EMBED Equation  so the required time shift would be  EMBED Equation  85. To get the time to reach the maximum height, we use Eq. 4-23, setting the left-hand side to zero. Thus, we find t =  EQ \f((60 m/s)sin(20o),9.8 m/s2)  = 2.094 s. Then (assuming a = 0) Eq. 10-13 gives q - qo = wo t = (90 rad/s)(2.094 s) = 188 rad, which is equivalent to roughly 30 rev. 86. In the calculation below, M1 and M2 are the ring masses, R1i and R2i are their inner radii, and R1o and R2o are their outer radii. Referring to item (b) in Table 10-2, we compute I =  EQ \f(1,2) M1 (R1i2 + R1o2) +  EQ \f(1,2) M2 (R2i2 + R2o2) = 0.00346  EMBED Equation.DSMT4  . Thus, with Eq. 10-38 (t = rF where r = R2o) and t = Ia (Eq. 10-45), we find a =  EQ \f((0.140)(12.0),0.00346)  = 485 rad/s2 . Then Eq. 10-12 gives w = at = 146 rad/s. 87. We choose positive coordinate directions so that each is accelerating positively, which will allow us to set abox = Ra (for simplicity, we denote this as a). Thus, we choose downhill positive for the m = 2.0 kg box and (as is conventional) counterclockwise for positive sense of wheel rotation. Applying Newtons second law to the box and (in the form of Eq. 10-45) to the wheel, respectively, we arrive at the following two equations (using q as the incline angle 20, not as the angular displacement of the wheel).  EMBED Equation  Since the problem gives a = 2.0 m/s2, the first equation gives the tension T = m (g sin q  a) = 2.7 N. Plugging this and R = 0.20 m into the second equation (along with the fact that a = a/R) we find the rotational inertia I = TR2/a = 0.054 kg EMBED Equation m2. 88. (a) We use t = Ia, where t is the net torque acting on the shell, I is the rotational inertia of the shell, and a is its angular acceleration. Therefore,  EMBED Equation  (b) The rotational inertia of the shell is given by I = (2/3) MR2 (see Table 10-2 of the text). This implies  EMBED Equation  89. Equation 10-40 leads to t = mgr = (70 kg) (9.8 m/s2) (0.20 m) = 1.4 102  EMBED Equation.DSMT4 . 90. (a) Equation 10-12 leads to  EMBED Equation  (b) Equation 10-15 leads to  EMBED Equation  (c) Dividing the previous result by 2p we obtain q = 39.8 rev. 91. We employ energy methods in this solution; thus, considerations of positive versus negative sense (regarding the rotation of the wheel) are not relevant. (a) The speed of the box is related to the angular speed of the wheel by v = Rw, so that  EMBED Equation  implies that the angular speed is w = 1.41/0.20 = 0.71 rad/s. Thus, the kinetic energy of rotation is  EMBED Equation  (b) Since it was released from rest at what we will consider to be the reference position for gravitational potential, then (with SI units understood) energy conservation requires  EMBED Equation.DSMT4  Therefore, h = 16.0/58.8 = 0.27 m. 92. (a) The time for one revolution is the circumference of the orbit divided by the speed v of the Sun: T = 2pR/v, where R is the radius of the orbit. We convert the radius:  EMBED Equation  where the ly  EMBED Equation  conversion can be found in Appendix D or figured  from basics (knowing the speed of light). Therefore, we obtain  EMBED Equation  (b) The number of revolutions N is the total time t divided by the time T for one revolution; that is, N = t/T. We convert the total time from years to seconds and obtain  EMBED Equation  93. The applied force P will cause the block to accelerate. In addition, it gives rise to a torque that causes the wheel to undergo angular acceleration. We take rightward to be positive for the block and clockwise negative for the wheel (as is conventional). With this convention, we note that the tangential acceleration of the wheel is of opposite sign from the blocks acceleration (which we simply denote as a); that is,  EMBED Equation.DSMT4 . Applying Newtons second law to the block leads to  EMBED Equation.DSMT4 , where T is the tension in the cord. Similarly, applying Newtons second law (for rotation) to the wheel leads to  EMBED Equation.DSMT4 . Noting that Ra = at =  a, we multiply this equation by R and obtain  EMBED Equation.DSMT4  Adding this to the above equation (for the block) leads to EMBED Equation  Thus, the angular acceleration is  EMBED Equation  With  EMBED Equation.DSMT4 , EMBED Equation.DSMT4  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4 , we find  EMBED Equation  or |a| = 4.6 rad/s2 , where the negative sign in a should not be mistaken for a deceleration (it simply indicates the clockwise sense to the motion). 94. (a) The linear speed at t = 15.0 s is  EMBED Equation  The radial (centripetal) acceleration at that moment is  EMBED Equation  Thus, the net acceleration has magnitude:  EMBED Equation  (b) We note that  EMBED Equation . Therefore, the angle between  EMBED Equation  and  EMBED Equation  is  EMBED Equation  so that the vector is pointing more toward the center of the track than in the direction of motion. 95. The distances from P to the particles are as follows:  EMBED Equation  The rotational inertia of the system about P is  EMBED Equation  which yields  EMBED Equation.DSMT4  for M = 0.40 kg, a = 0.30 m, and b = 0.50 m. Applying Eq. 10-52, we find  EMBED Equation  96. In the figure below, we show a pull tab of a beverage can. Since the tab is pivoted, when pulling on one end upward with a force  EMBED Equation.DSMT4 , a force  EMBED Equation.DSMT4  will be exerted on the other end. The torque produced by  EMBED Equation.DSMT4  must be balanced by the torque produced by  EMBED Equation.DSMT4  so that the tab does not rotate.  The two forces are related by  EMBED Equation.DSMT4  where  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . Thus, if F1 = 10 N,  EMBED Equation.DSMT4  97. The centripetal acceleration at a point P that is r away from the axis of rotation is given by Eq. 10-23:  EMBED Equation.DSMT4 , where  EMBED Equation.DSMT4 , with EMBED Equation.DSMT4  (a) If points A and P are at a radial distance rA = 1.50 m and r = 0.150 m from the axis, the difference in their acceleration is  EMBED Equation.DSMT4 . (b) The slope is given by  EMBED Equation.DSMT4 . 98. Let T be the tension on the rope. From Newtons second law, we have  EMBED Equation.DSMT4 . Since the box has an upward acceleration a = 0.80 m/s2, the tension is given by  EMBED Equation.DSMT4  The rotation of the device is described by  EMBED Equation.DSMT4 . The moment of inertia can then be obtained as  EMBED Equation.DSMT4  99. (a) With r = 0.780 m, the rotational inertia is  EMBED Equation  (b) The torque that must be applied to counteract the effect of the drag is  EMBED Equation  100. We make use of Table 10-2(e) as well as the parallel-axis theorem, Eq. 10-34, where needed. We use  EMBED Equation  (as a subscript) to refer to the long rod and s to refer to the short rod. (a) The rotational inertia is  EMBED Equation  (b) We note that the center of the short rod is a distance of h = 0.25 m from the axis. The rotational inertia is  EMBED Equation  which again yields I = 0.019 kg EMBED Equation m2. 101. (a) The linear speed of a point on belt 1 is  EMBED Equation.DSMT4 . (b) The angular speed of pulley B is  EMBED Equation.DSMT4 . (c) Since the two pulleys are rigidly attached to each other, the angular speed of pulley  EMBED Equation.DSMT4  is the same as that of pulley B, that is,  EMBED Equation.DSMT4 . (d) The linear speed of a point on belt 2 is  EMBED Equation.DSMT4 . (e) The angular speed of pulley C is  EMBED Equation.DSMT4  102. (a) The rotational inertia relative to the specified axis is  EMBED Equation  which is found to be I = 4.6 kg EMBED Equation m2. Then, with w = 1.2 rad/s, we obtain the kinetic energy from Eq. 10-34:  EMBED Equation  (b) In this case the axis of rotation would appear as a standard y axis with origin at P. Each of the 2M balls are a distance of r = L cos 30 from that axis. Thus, the rotational inertia in this case is  EMBED Equation  which is found to be I = 4.0 kg EMBED Equation m2. Again, from Eq. 10-34 we obtain the kinetic energy  EMBED Equation  103. We make use of Table 10-2(e) and the parallel-axis theorem in Eq. 10-36. (a) The moment of inertia is  EMBED Equation.DSMT4  (b) The rotational kinetic energy is  EMBED Equation.DSMT4 . The linear speed of the end B is given by  EMBED Equation.DSMT4 , where rAB is the distance between A and B. (c) The maximum angle q is attained when all the rotational kinetic energy is transformed into potential energy. Moving from the vertical position (q = 0) to the maximum angle q , the center of mass is elevated by  EMBED Equation.DSMT4 , where dAC = 1.00 m is the distance between A and the center of mass of the rod. Thus, the change in potential energy is  EMBED Equation.DSMT4  which yields  EMBED Equation.DSMT4 , or  EMBED Equation.DSMT4 . 104. (a) The particle at A has r = 0 with respect to the axis of rotation. The particle at B is r = L = 0.50 m from the axis; similarly for the particle directly above A in the figure. The particle diagonally opposite A is a distance  EMBED Equation  from the axis. Therefore,  EMBED Equation  (b) One imagines rotating the figure (about point A) clockwise by 90 and noting that the center of mass has fallen a distance equal to L as a result. If we let our reference position for gravitational potential be the height of the center of mass at the instant AB swings through vertical orientation, then  EMBED Equation.DSMT4  Since h0 = L = 0.50 m, we find K = 3.9 J. 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FMathType 5.0 Equation MathType EFEquation.DSMT49q w avg =(2.5)(2p)1.4=11 rad/s.<<<<ObjInfoOle10NativeprEquation Native _1106308083u F@n`@n`A|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  w avg == (2.5 rev)(2p rad/rev)1.4s==11 rad/s.Ole CompObjtv\ObjInfowOle10Native FMathType Equation Equation Equation9q q=4.0 rad/s()t-3.0 rad/s 2 ()t 2 +1.0 rad/s 3 ()t 3 FMathType Equation Equation Equation9q w=ddt4t-3t 2 +t 3 ()=4-6t+3t 2 ._1106308084s}z F0Qn`0Qn`Ole CompObjy{\ObjInfo|Ole10Native_1106308085 F0n`0n`Ole CompObj~\ FMathType Equation Equation Equation9q a avg =w 4 -w 2 4-2=12 rad/s 2 .ObjInfoOle10Native_1106308086 F n` n`Ole  FMathType Equation Equation Equation9q a=dwdt=ddt4-6t+3t 2 ()=-6+6t.CompObj\ObjInfoOle10Native_1106308320 F }n` }n`Ole CompObj\ObjInfoOle10Native FMathType Equation Equation Equation9q Dt=1/8 rev2.5 rev/s=0.050 s.  &_1106308321 F }n` }n`Ole CompObj\ObjInfo    !#$%&'(+./023456789<?@ABCDEFILMOPQRSTUX[\]^_`abcfijkmnopqrstwz{|} FMathType Equation Equation Equation9q v min =20 cm0.050 s=400 cm/s=4.0 m/s.Ole10Native_1132342895Fn`n`Ole CompObji FMathType 5.0 Equation MathType EFEquation.DSMT49q «dDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  25 rad== 12a(5.0 s) ObjInfo Equation Native  _1106308430 Fn`n`Ole 2 !a==2.0 rad/s 2 . FMathType Equation Equation Equation9q w avg =DqDt=CompObj\ObjInfoOle10Native_1214774900DFdn`dn`25 rad5.0 s=5.0 rad/s.~`~`~`~` FMathType 5.0 Equation MathType EFEquation.DSMT49q w=2.0 rad/s 2 Ole CompObjiObjInfoOle10Native()(5.0s)=10 rad/s .DzHA˜ DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  w==2.0 rad/s 2 ()(5.0s)==Equation Native "_1214769186Fdn`dn`Ole )CompObj*i10 rad/s . FMathType 5.0 Equation MathType EFEquation.DSMT49q q=w 0 +12at 2 =0+12(2.0)(10) 2 =100rad. 8,> ObjInfo,Ole10Native-Equation Native 1&_1214760667Fn`n`DDA |DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  q==w 0 ++ 12at 2 ==0++ 12(2.0 rad/s 2 )(10s) 2 ==100rad. FMathType 5.0 Equation MathType EFEquation.DSMT49qA|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@Ole :CompObj;iObjInfo=Equation Native >AE%B_AC_AE*_FA@AHA*_D_E_E_A  w==w 0 ++at!t== 0"-120 rad/s"-4.0 rad/s 2 ==30 s._1214760713~Fn`n`Ole GCompObjHiObjInfoJ FMathType 5.0 Equation MathType EFEquation.DSMT49q q=12(w 0 +w)t=12(120+0)(30)A|DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AOle10NativeKEquation Native N_1214760914Fn`n`Ole VE%B_AC_AE*_FAGAHA*_D_E_E_A  q== 12(w 0 ++w)t== 12(120 rad/s++0)(30s)==1.810 3  rad. FMathType 5.0 Equation MathTyCompObjWiObjInfoYEquation Native Zk_1214760913Fn`n`pe EFEquation.DSMT49qAO|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  w==w 0 ++at!a== (3000"-1200) rev/min(12/60) min==9.010 3  rev/min 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49qOle dCompObjeiObjInfogOle10Nativeh q=12(w 0 +w)t=12(1200+3000)1260()A|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  q== Equation Native l;_1108934547FJn`Jn`Ole uCompObjvi12(w 0 ++w)t== 12(1200 rev/min++3000 rev/min) 1260 min() FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfoxEquation Native y_1132343010IFn`n`Ole ~|XDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT ExtraWinAllCodePagessө!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  4.210 2 FMathType 5.0 Equation MathTyCompObjiObjInfoEquation Native _1214769720F,n`,n`pe EFEquation.DSMT49q <DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  w 2 ==0 FMathType 5.0 Equation MathTyOle CompObjiObjInfoOle10Nativepe EFEquation.DSMT49q q 2 =12w 0 +w 2 ()t 2 t 2 =2(40)0.239ϟϟϟϟϟϟϟϟ????A4|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT ExtEquation Native P_1108935088Fn`n`Ole CompObjira!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  q 2 == 12w 0 ++w 2 ()t 2 !t 2 == 2(40 rev)0.239 rev/s==335s FMathType 5.0 Equation MathType EFEquation.DSMT49q+lXDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT ExtraWinAllCodePagessө!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 2ObjInfoEquation Native G_1214769894:Fn`n`Ole  H"3.410 2  s FMathType 5.0 Equation MathType EFEquation.DSMT49q q 2 =w 2 t 2 -12at 22 a=-2(    `"!A$#&%'()*+,-./0123456789:;<=>?@BDaCFEGHIJKLMONRPQSTUVWXY[Z\]^cbabcdefghijklmnopqrstuvwxyz{|}~j~W.Uvb"̇o6 -11WwA(Zƺ۔MڬfZ(@E `4P@I.UlEu:F<=da׬8Ǎ:[zL*yLuDd\l\  c $A? ?#" `2hmo8b`!hmo8b` 0gxڥSkA3횚C覠BdLolӨH1)7Y4/m.~!RO?*xXZUHٓ{T ao޼y}`@9;@#3 Ƈál>Н#08 SX!C qƩXdVm`1^ \BqOs98] iD@RorݴтWl|T0229k]嗧{71?Ʃ&jVt803qx;pw UgŞ,,~Vv[a*͎zę6Ko3We.!@}c[_O[-*Y쮛hVv]3׌;$1'*uf!&*Jq5`zbH]յsg{&[tϻQ4RePt+q}p E>-fH8u XHO> KOs w$TU8I4,G . 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FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoOle10NativeEquation Native  w 1 =w 0 +at 1 t 1 =101.04=9.6 s.A|DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  w 1 ==     #$%(+,-.145789:;<?BCDEHKLMPSTVWXYZ[\]`cdefilmnoruvwx{~w 0 ++at 1 !t 1 == 10 rev/s1.04 rev/s 2 ==9.6 s. FMathType 5.0 Equation MathType EFEquation.DSMT49q_1214769579F o` o`Ole CompObjiObjInfoOle10NativeEquation Native  :_1318021395FD o`D o`Ole  w 12 =w 02 +2aq 1 q 1 =10 2 2(1.04)=4.8 rev.cA|DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  w 12 ==w 02 ++2aq 1 !q 1 == (10 rev/s) 2 2(1.04 rev/s 2 )==48 rev. FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoEquation Native _1318021420Fo`o`ElDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  q==  12 at 2 FMathType 5.0 Equation MathTyOle CompObjiObjInfo!Equation Native "pe EFEquation.DSMT49qElDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 1 FMathType 5.0 Equation MathTy_1318021452F&o`&o`Ole &CompObj'iObjInfo)pe EFEquation.DSMT49qE DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 2 ==t 1 ++4.0sEquation Native *)_1318021521 Fpo`po`Ole /CompObj  0i FMathType 5.0 Equation MathType EFEquation.DSMT49q q=w 0 t+ 12 at 2E{DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA ObjInfo2Ole10Native 3Equation Native 6_1318021608 Fo`o`AE%B_AC_AE*_FAGAHA*_D_E_E_A  q 1 == 12at 12 ,q 2 == 12at 22 FMathType 5.0 Equation MathType EFEquation.DSMT49qOle =CompObj>iObjInfo@Equation Native AE$DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  Dq==q 2 "-q 1 FMathType 5.0 Equation MathTy_1318021639FRo`Ro`Ole FCompObjGiObjInfoIpe EFEquation.DSMT49qEDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 1 FMathType 5.0 Equation MathTyEquation Native J_1318021789$Fpo`po`Ole NCompObjOipe EFEquation.DSMT49q q=w 0 t+ 12 at 2E|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A ObjInfoQOle10NativeREquation Native U6_1318021877!Fp o`p o` Dq==q 2 "-q 1 == 12at 22 "-t 12 ()== 12a(t 2 ++t 1 )(t 2 "-t 1 ) FMathType 5.0 Equation MathTyOle ^CompObj "_iObjInfo#aEquation Native b pe EFEquation.DSMT49qEDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  Dq==120rad_1318021941)&F`Wo``Wo`Ole gCompObj%'hiObjInfo(j FMathType 5.0 Equation MathType EFEquation.DSMT49qE|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  a==3.0rad/s 2Equation Native k _1318021975+F`!o``!o`Ole pCompObj*,qi FMathType 5.0 Equation MathType EFEquation.DSMT49qE lDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 2 "-t ObjInfo-sEquation Native t)_1318022008C0F`9$o``9$o`Ole y1 ==4.0s FMathType 5.0 Equation MathType EFEquation.DSMT49q q=w 0 t+ 12 at 2CompObj/2ziObjInfo|Ole10Native13}Equation Native EE)lDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 2 ++t 1 == 2(Dq)a(t 2 "-t 1 )== 2(120rad)(3.0rad/s 2 )(4.0s)==20s FMathType 5.0 Equation MathType EFEquation.DSMT49qEDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA _13180221506FP&o`P&o`Ole CompObj57iObjInfo8Equation Native _13180221664>;FP(o`P(o`Ole CompObj:<iAE%B_AC_AE*_FAGAHA*_D_E_E_A  t 2 ==12.0s FMathType 5.0 Equation MathType EFEquation.DSMT49qEDSMT5WinAllBasicCodePagesObjInfo=Equation Native  _1318022298@FPe+o`Pe+o`Ole Times New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  t 1 ==8.0s FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObj?AiObjInfoBEquation Native _13180223089EF@-o`@-o`E,DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  q 1 FMathType 5.0 Equation MathType EFEquation.DSMT49qOle CompObjDFiObjInfoGEquation Native EDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  q 2 FMathType 5.0 Equation MathType EFEquation.DSMT49q_1325202955!JF0j2o`0j2o`Ole CompObjILiObjInfoOle10NativeKMEquation Native _1318022457PF04o`04o`Ole  q=w 0 t+ 12 at 2lDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  q 1 == 12at 12 == 12(3.0rad/s 2 )(8.0s) 2 ==96radq 2 == 12at 22 == 12(3.0rad/s 2 )(12.0s) 2 ==216rad. 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LEDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  w 0 == (200rev/min)(2prad/rev)60s/min==20.9rad/s. FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoEquation Native _1106309152 Fp~o`p~o`E‡DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  v==rw 0 ==(0.60 m)(20.9 rad/s)==12.5m/s.Ole CompObj\ObjInfoOle10Native FMathType Equation Equation Equation9q a=w-w o t=800rev/min 2 . FMathType 5.0 Equation MathTy_1220205112 5Fp&o`p&o`Ole CompObjiObjInfope EFEquation.DSMT49q q=12w o +w()t=12(200+100)(1)=600rev.y-|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT ExtOle10NativeEquation Native I_1318023495Fpo`po`Ole ra!/ED/APG_APA AE%B_AC_AE*_FA@AHA*_D_E_E_A  q== 12w o ++w()t== 12(200 rev/min++1000 rev/min)(1.0 min)==600rev. FMathType 5.0 Equation MathType EFEquation.DSMT49q q=12w o +w()t=12(200+100)(1)=600rev.CompObjiObjInfoOle10NativeEquation Native E°lDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA AE%B_AC_AE*_FA@AHA*_D_E_E_A  q==q 0 ++w 0 t++ 12at 2 ==0++(200 rev/min)(1.0 min)++ 12(800 rev/min 2 )(1.0 min) 2 ==600rev. 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"Y22x}>zW^}U`!u2x}>zW^)@CxڭUOSA-XJc[>LI _$H(P RcК~H[M4z2=x0Ą$L<Ic ۶B?x~;_ Ķ0A7pRIx+~V%\4ڠJ `G&XehZ8CE0uM(Do8i- Қ%Jt7|҇?i3`'PBô2? X-Qa=-p+KOG-\pv,>6K[h%0xxñQ.9r*M¯b|o*$G=οqY0>߼dc1"8^%.zk.UHzѶk ܆g6?`k3zH6F~F\\r* ~X]! i,P" cP~P$*^'M2}@$[RP]n$غUQU>]mmt&~ɞqG}7[^° {nEX'2׆"p) ~X,:N|,3`Be $hm}Pa l({l'KC:m gӺ5Tp0Kw^RbX aYV<. }*T >6hObUx"k>c{uSDt7Ak60▫(ŲBbFbDDDD <t>L&,|㹡Gi %=] y_A?yLRoUR}chDd\ t c $As? ?#" `[2-$jc1G6Ku `!$jc1G6Ku ``P!xuRAKAFHA) XK=Ȫk{I Mꢁ& ^#BAD"{+4}3(8y}of.\,y6ؾ1INc/$ًV ޟ #=IY_?j,Qh\S/jPF51Z̗A}RYvڎWpw:ұnObjInfoEquation Native N_1108940181-F@ȑo`@ȑo`Ole A2DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A   1Dt== wrdH"199/s FMathType 5.0 Equation MathTyCompObjiObjInfoOle10NativeEquation Native pe EFEquation.DSMT49q w=2Symbolprad8.6410 4 s=7.2710 -5 rad/s.lXDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT ExtraWinAllCodePagessө!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  w== 2prad8.6410 4 s==7.310 "-5 rad/s. FMathType 5.0 Equation MathTy_1132422469F@9o`@9o`Ole CompObjiObjInfo!pe EFEquation.DSMT49q v=w(Rcos40)=(7.2710 -5 rad/s)(6.3710 6 m)cos40=355m/s.>%6<DSMT5WinAllBasicCodePagesOle10Native"Equation Native 'R_1132422500F@o`@o`Ole 1Times New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  v==w(Rcos40)==(7.310 "-5 rad/s)(6.410 6 m)cos40==3.510 2 m/s. FMathType 5.0 Equation MathType EFEquation.DSMT49q v=wR=(7.2710 -5 rad/s)(6.3710 6 m)=463m/s.Times New CompObj2iObjInfo4Ole10Native5Equation Native 9Roma%<DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  v==wR==(7.310 "-5 rad/s)(6.410 6 m)==4.610 2 m/s. FMathType 5.0 Equation MathType EFEquation.DSMT49qE DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A _1318024718 F0o`0o`Ole ACompObj  BiObjInfo DEquation Native E_1106311559 F0eo`0eo`Ole JCompObjK\ r==Rcos90==0 FMathType Equation Equation Equation9q a=DwDt=0-150 rev/min(2.2 h)(60 miObjInfoMOle10NativeN_1132422572F0֝o`0֝o`Ole Sn/1h)=-1.14 rev/min 2 .ǀǀǐǐǀǀ FMathType 5.0 Equation MathType EFEquation.DSMT49q q=w 0 t+12atCompObjTiObjInfoVOle10NativeWEquation Native ^ 2 =(150 rev/min)(132 min)+12-1.14 rev/min 2 ()132 min() 2 =9.910 3 rev.%´<DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  q==w 0 t++ 12at 2 ==(150 rev/min)(132 min)++ 12"-1.14 rev/min 2 ()132min() 2 ==9.910 3 rev. FMathType Equation Equation Equation9q@ a t =ar=-1.14rev/min 2 ()2_1106311562a) F o` o`Ole jCompObjk\ObjInfom&      !"#$'%(F)*+-,/.0123546798:;<=>@?ABDCEIGHKJLNMOPQRSUTXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Ole10NativenD_1214775537#Fۤo`ۤo`Ole tCompObj!uiSymbolprad1rev()1min60s() 2 (500mm)oldH FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfowOle10Native "x$Equation Native }_1214775560%FLo`Lo`  a r =w 2 r=(75rev/min)2Symbolprad/rev1min/60s()() 2 (0.50m)eHelveticADSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  w==(75rev/min)(2prad/rev)(1 min/ 60 s)==7.85 rad/s. FMathType 5.0 Equation MathType EFEquation.DSMT49qOle CompObj$'iObjInfoOle10Native&($  a r =w 2 r=(75rev/min)2Symbolprad/rev1min/60s()() 2 (0.50m)eHelveticADSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT ExtEquation Native _1106311564+ Fo`o`Ole CompObj*,\ra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  a r ==w 2 r==(7.85 rad/s) 2 (0.50m)H"31m/s 2 FMathType Equation Equation Equation9q |2a |= a r2 +a t2 a r =31m/s 2 .bG'v G' FMathType Equation Equation ObjInfo-Ole10Native_1106311625yB0 Fxo`xo`Ole CompObj/1\ObjInfo2Ole10Native_11063116275 Fo`o`Equation9q w=3313rev/min()2Symbolprad/rev60s/min()=3.49rad/s.3Ole CompObj46\ObjInfo7Ole10Native FMathType Equation Equation Equation9q a=w 2 r=3.49rad/s() 2 (6.010 -2 m)=0.73m/s 2 ._11063116283=: Fo`o`Ole CompObj9;\ObjInfo< FMathType Equation Equation Equation9q m s,min =ag=0.739.8=0.075.``` FMathType Equation Equation Ole10Native_1106311629? Fo`o`Ole CompObj>@\Equation9q |2a |= a r2 +a t2 = (w 2 r) 2 +(ar) 2 =r w 4 +a 2 .ObjInfoAOle10Native_11063116308D Fo`o`Ole CompObjCE\ObjInfoFOle10Native_1132422620IF^o`^o` FMathType Equation Equation Equation9q f s,max =m s mg=ma max =mr w max4 +a 2 .yOle CompObjHKiObjInfoOle10NativeJL FMathType 5.0 Equation MathType EFEquation.DSMT49q m s,min =r w max4 +a 2 g=r w max4 +(w max /t) 2 g=(0.060) 3.49 4 +(3.4/0.25) 2 9.8=0.11.%T<,DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  m s,min == r w max4 ++a 2 g== r Equation Native _1106309416|O Fϼo`ϼo`Ole CompObjNP\w max4 ++(w max /t) 2 g== (0.060) 3.49 4 ++(3.4/0.25) 2 9.8==0.11. FMathType Equation Equation Equation9q@ a C =r A r C ()a C =10cm25cm()(1.6rad/s 2 )=0.64rad/s 2 .ObjInfoQOle10NativeD_1322420516TFЊo`Њo`Ole  FMathType 5.0 Equation MathType EFEquation.DSMT49qlDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  w C ==a CompObjSUiObjInfoVEquation Native _1106309418qY FEo`Eo`C t FMathType Equation Equation Equation9q t=w C a C =10.5rad/s0.64rad/s 2 =16s.NOle CompObjXZ\ObjInfo[Ole10Native_ FMathType Equation Equation Equation9q t=2lc=2(500m)2.99810 8 m/s=3.3410 -6 s_1106309440^ FEo`Eo`Ole CompObj]_\ObjInfo`Ole10Native_1106309442\ c Fo`o`Ole CompObjbd\   !"%()*+.12356789:;<?BCDGJKLORSVYZ[^abcdgjklmnopqrsvyz{|} FMathType Equation Equation Equation9q w=qt=1.2610 -2 rad3.3410 -6 s=3.810 3 radObjInfoeOle10Native_1108940492hFo`o`Ole  /s.8<<; FMathType 5.0 Equation MathType EFEquation.DSMT49q v=wr=3.810 3 rad/s()0.05m()=1CompObjgj iObjInfo Ole10NativeikEquation Native  90m/s. lXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  v==wr==3.810 3 rad/s()0.050m()==1.910 2 m/s. FMathType Equation Equation Equation9q a t =ar=14.2_1106309356n Fo`o`Ole CompObjmo\ObjInfopOle10Native_1106309358lMs Fo`o`Ole #CompObjrt$\rad/s 2 ()(2.83cm)=40.2cm/s 2 . FMathType Equation Equation Equation9q a r =w 2 r=(289ObjInfou&Ole10Native'_1220205258~xF,o`,o`Ole ,rad/s) 2 (0.0283m)=2.3610 3 m/s 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjwz-iObjInfo/Ole10Nativey{0Equation Native 44 q=w 2 2a=289 2 2(14.2)=2.9410 3 rad.3y|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA AE%B_AC_AE*_FA@AHA*_D_E_E_A  q== w 2 2a== (289 rad/s) 2 2(14.2 rad/s 2 )==2.9410 3 rad. FMathType Equation Equation Equation9q_1106309360~ Fo`o`Ole =CompObj}>\ObjInfo@ s=rq=(0.0283m)(2.9410 3 rad)=83.2m.P  FMathType 5.0 Equation MathType EFEquation.DSMT49qOle10NativeA_1133200393bjFXo`Xo`Ole ECompObjFiObjInfoHEquation Native I_1106311676 Fo`o`Ole M-TD,DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  ! FMathType Equation Equation Equation9qCompObjN\ObjInfoPOle10NativeQ_1106311678 Fo`o` a=dwdt=-2SymbolpT 2 dTdt.???? FMathType Equation Equation Equation9qOle TCompObjU\ObjInfoWOle10NativeX dTdt=1.2610 -5 s/y3.1610 7 s/y=4.0010 -13 . FMathType Equation Equation Equation9q_1106311679 Fo`o`Ole \CompObj]\ObjInfo_Ole10Native`,_1132422950GFp?o`p?o`Ole eCompObjfi( a=-2Symbolp(0.033s) 2 ()(4.0010 -13 )=-2.310 -9 rad/s 2 .\N\N FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfohEquation Native i_1106311681XF`ko``ko`Ole t%—tDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  t=="- w 0 a=="- 2paT=="- 2p("-2.310 "-9 rad/s 2 )(0.033s)==8.310 10 sH"2.610 3   years FMathType 4.0 Equation MathType EFEquation.DSMT49qCompObjuiObjInfowOle10NativexDEquation Native ~@ w=w 0 +at+2SymbolpT+at2p0.033s+(-2.310 9 rad/s 2 )(-2.9610 10 s)=258rad/s.ט@6M6GxDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  w==w 0 ++at++ 2pT++at== 2p0.033s++("-2.310 "-9 rad/s 2 )("-2.9610 10 s)==258rad/s. FMathType Equation Equation Equation9q T=2Symbolpw=2p258rad/s=2.410 -2 s.+v +_1106311682 F`o``o`Ole CompObj\ObjInfoOle10Native_1106315590 FP&o`P&o`Ole CompObj\ FMathType Equation Equation Equation9q` K= 12 Iw 2 ,ObjInfoOle10Natived_1106315591 FPo`Po`Ole CompObj\ObjInfoOle10Nativel_1106315592 FPo`@Ro` FMathType Equation Equation Equation9qh kgm 2Q%DO.\I\I FMathType Equation Equation Equation9qOle CompObj\ObjInfoOle10Native w=(602rev/min)(2Symbolprad/rev)60s/min=63.0rad/s.j v ,橁 FMathType Equation Equation Equation9q_1106315593 F@Ro`@Ro`Ole CompObj\ObjInfo I=2Kw 2 =2(24400J)(63.0rad/s) 2 =12.3kgm 2 .EKJ FMathType Equation Equation Equation9qOle10Native_1106315999 F@o`@o`Ole CompObj\ObjInfoOle10Natived_1106316000F@4o`@4o`Ole ` I= 12 MR 2 FMathType 4.0 Equation MathType EFEquation.DSMT49q K-12Iw 2 =1%nW;}]a$0{dZU1eI NeQ*"M眵<9?0U嶒 I64$Ύ ,B?v=N'61훚A}Yl^U dzMo矔jDdlJ u C At? 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FMathType 5.0 Equation MathType EFEquation.DSMT49q K= 14 (1.25)(0.25) 2 (235) 2 =1.1CompObjiObjInfoOle10NativeEquation Native O10 3 J.3lDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  K 2 == 14(1.25kg)(0.75 m) 2 (235rad/s) 2 ==9.7110 3 JH"9.710 3 J. FMathType 5.0 Equation MathType EFEquation.DSMT49q_1325203324F 9p` p`Ole CompObjiObjInfoEquation Native _1106316158 F  p`  p`Ole CompObj\DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  kg"m 2 FMathType Equation Equation Equation9qObjInfoOle10Native$_1106316160 F  p`  p`Ole   I com =112ML 2 =1120.56kg()1.0m() 2 =4.6710 -2 kgm 2 .0 FMathType Equation Equation Equation9qCompObj\ObjInfoOle10Native$_1132439589Fe p`e p`  I=4.6710 -2 kgm 2 +0.56kg()0.30m() 2 =9.710 -2 kgm 2 .  FMathType 5.0 Equation MathTyOle CompObjiObjInfoOle10Natived      !"#$%&')*+,-./0123456789:;>ABCDFGHIJKLMNOPQRUXYZ[\]_`abcdefghijmpqrstux{~pe EFEquation.DSMT49q` K=12Iw 2 =1212MR 2 ()w 2 =14(500kg)(200p rad/s) 2 (1.0m) 2 =4.910 7 J. 2 hLEquation Native _1106323705 FGp`Gp`Ole CompObj\%{tDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  K== 12Iw 2 == 12 12MR 2 ()w 2 == 14(500kg)(200p rad/s) 2 (1.0m) 2 ==4.910 7 J. FMathType Equation Equation Equation9q t=KP=4.910 ObjInfoOle10Native_1220205363Fp`p`Ole 7 J8.010 3 W=6.210 3 s FMathType 5.0 Equation MathType EFEquation.DSMT49q  I=I 1 +I 2CompObjiObjInfoOle10Native$Equation Native ( +I 3 +I 4 =112Md 2 +M12d() 2 ()+md 2 +112Md 2 +M32d() 2 ()+m(2d) 2 =83Md 2 +5md 2 .y°|DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I==I 1 ++I 2 ++I 3 ++I 4 == 112Md 2 ++M 12d() 2 ()++md 2 ++ 112Md 2 ++M 32d() 2 ()++m(2d) 2 == 83Md 2 ++5md 2 == 83(1.2 kg)(0.056 m) 2 +5(0.85 kg)(0.056 m) 2 =0.023 kg"m 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49q_1132439331Fp`p`Ole <CompObj=iObjInfo?Ole10Native@Equation Native E[_1220205606Fp`p`Ole S K=12Iw 2 =43Md 2 +52md 2 ()w 2 .L i%?tDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  K== 12Iw 2 == 43M++ 52m()d 2 w 2 == 43(1.2 kg)++ 52(0.85 kg)[](0.056 m) 2 (0.30 rad/s) 2 ==1.110 "-3  J. FMathType 5.0 Equation MathType EFEquation.DSMT49q I x =m i y iCompObjTiObjInfoVOle10NativeWEquation Native ^?2 =50 i=14  2.0() 2 +25() 4.0() 2 +25 -3.0() 2 +304.0() 2 =1.310 3  gcm 2 .y#|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I x ==pm i y i2 == i==14 " 502.0() 2 ++25()4.0() 2 ++25"-3.0() 2 ++304.0() 2 []g"cm 2 ==1.310 3  g"cm 2 . FMathType Equation Equation Equation9q I y =m i x i_1106323470  F-p`-p`Ole kCompObjl\ObjInfonOle10Nativeo_1106323471 Fwp`wp`Ole vCompObjw\2 =50 i=14  2.0() 2 +25() 0() 2 +25 3.0() 2 +302.0() 2 =5.510 2  gcm 2 . FMathType Equation Equation Equation9q`  x 2 +y 2 FMathType Equation Equation Equation9qObjInfoyOle10Nativezd_1106323472 F p` p`Ole |CompObj }\ObjInfo Ole10Natived_1106316311  FY#p`Y#p`` I z =m i x i2 +y i2 ()=I x +I y = i=14  1.310 3 +5.510 2 =1.910 2  gcm 2 .+>  FMathType Equation Equation Ole CompObj \ObjInfoOle10NativeEquation9q I com =M12a 2 +b 2 (). FMathType Equation Equation Equation9q_1106316312; FУ%p`У%p`Ole CompObj\ObjInfo h= a/2() 2 +b/2() 2 FMathType 5.0 Equation MathType EFEquation.DSMT49qOle10Native_1318028608CF(p`(p`Ole CompObjiObjInfoOle10NativeHEquation Native X_1318028509%F^*p`^*p`D I=I com +Mh 2 =M12 a 2 +b 2 ()+M4 a 2 +b 2 ()=M3 a 2 +b 2 ().ȌKȌKE<DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I==I com ++Mh 2 == M12a 2 ++b 2 ()++ M4a 2 ++b 2 ()== M3a 2 ++b 2 ().Ole CompObjiObjInfoEquation Native   FMathType 5.0 Equation MathType EFEquation.DSMT49qE,DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  M==0.172kg FMathType 5.0 Equation MathType EFEquation.DSMT49qDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A _1325203551H"F,p`,p`Ole CompObj!#iObjInfo$Equation Native _1318028544'F@/p`@/p`Ole CompObj&(i a==3.5 cm, FMathType 5.0 Equation MathType EFEquation.DSMT49qEDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA ObjInfo)Equation Native _1318028628,F3p`3p`Ole AE%B_AC_AE*_FAGAHA*_D_E_E_A  b==8.4 cm FMathType 5.0 Equation MathType EFEquation.DSMT49qD I=I com +Mh 2 =M12 a 2 +b 2 CompObj+.iObjInfoOle10Native-/HEquation Native u()+M4 a 2 +b 2 ()=M3 a 2 +b 2 ().ȌKȌKEYDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I== M3a 2 ++b 2 ()== 0.172 kg3[(0.035 m) 2 ++(0.084 m) 2 ]==4.710 "-4  kg"m 2 . 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FMathType Equation Equation Equation9q@ 2F  1%2_1106323712pQN FpKp`pKp`Ole CompObjMO\ObjInfoPOle10NativeD_1106323713S Fp:Np``Pp`Ole CompObjRT\ FMathType Equation Equation Equation9q@ 2F  2 FMathType Equation Equation Equation9qObjInfoUOle10NativeD_1106323714LX F`Rp``Rp`Ole CompObjWY\ObjInfoZ Ole10Native!_1132439835`]F`fUp``fUp` t=r 1 F 1 sinq 1 -r 2 F 2 sinq 2 .?___ FMathType 5.0 Equation MathType EFEquation.DSMT49qOle %CompObj\^&iObjInfo_(Equation Native )%Td,DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  t==(1.30m)(4.20N)sin75"-(2.15m)(4.90N)sin60=="-3.85N"m. FMathType 5.0 Equation MathType EFEquation.DSMT49q%^TDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A _1132439918bFP!Zp`P!Zp`Ole 1CompObjac2iObjInfod4Equation Native 5z_1106323706g FP\p`P\p`Ole CCompObjfhD\ t==t A ++t B ++t C ==F A r A sinf A "-F B r B sinf B ++F C r C sinf C ==(10)(8.0)sin135"-(16)(4.0)sin90++(19)(3.0)sin160==12N"m. 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"2i6*ŅdgEb|`!=6*Ņdgl` @| xڕRkQmRMh6B!lmcmMhNxYxԃ'A'u˦H}of{1} @KD1"EBrӼTэ9Kd2r#w| & =:r]%% ڢ%FD1RkRF kwsUkYM=C xD5y^%,, |WF⸂NÓC'Y\Pީ{~\9 ]Ozi,JW׷<ȟM z+DdP  S A? "25qrSI}unco`![5qObjInfoOle10NativeEquation Native W_1214774350Fp`p`l a,A;DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  m 1 g"-T 1 =="-m 1 a,Ole CompObjiObjInfoOle10Native$ FMathType 5.0 Equation MathType EFEquation.DSMT49q  T l =m l (g+a)=(0.460kg)9.8m/s 2 +6.0010 -2 m/s 2 ()=4.54N.   "#$%&'()*+.12345789:;<=>?@CFILMNOPQRUXYZ]`abehijmpqruxyz}A4DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  T 1 ==m 1 (g++a)==(0.460kg)9.8m/s 2 ++6.Equation Native P_1106323992 FϜp`Ϝp`Ole  CompObj \0010 "-2 m/s 2 ()==4.54N. FMathType Equation Equation Equation9q a=aR=6.0010ObjInfoOle10Native_1214774400F@p`@p`Ole  -2 m/s 2 5.0010 -2 m=1.20rad/s 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoEquation Native @_1214774444Fp`p`A$ DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  t==(T 2 "-T 1 )R FMathType 5.0 Equation MathType EFEquation.DSMT49qOle CompObjiObjInfo Equation Native !A˜ DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I== T 2 "-T 1 ()Ra== 4.87N"-4.54N()5.0010 "-2 m()1.20rad/s 2 ==1.3810 "-2 kg"m 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49q_1132440178Fp`p`Ole ,CompObj-iObjInfo/Ole10Native0dEquation Native 6_1106323930 Flp`lp`Ole A` t net =F 1 R 2 -F 2 R 2 -F 3 R 1 =(6.0N)(0.12m)-(4.0N)(0.12m)-(2.0N)(0.05m)=71Nm.Ti%ƒLDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  t net ==F 1 R"-F 2 R"-F 3 r==(6.0N)(0.12m)"-(4.0N)(0.12m)"-(2.0N)(0.050m)==71N"m. FMathType Equation Equation Equation9q` I= 12 MR 22P&NCompObjB\ObjInfoDOle10NativeEd_1133200846Fݨp`ݨp`Ole GCompObjHiObjInfoJEquation Native K FMathType 5.0 Equation MathType EFEquation.DSMT49q-DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  a== t net I== 71N"m  12 (2.0kg)(0.12 m) 2 ==9.7rad/s 2 FMathType 5.0 Equation MathType EFEquation.DSMT49q%LDSMT5WinAllBasicCodePages_1132440102[,Fp`p`Ole SCompObjTiObjInfoVEquation Native W_1325203653Fpp`pp`Ole [CompObj\iTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  a==w/t FMathType 5.0 Equation MathType EFEquation.DSMT49qlDSMT5WinAllBasicCodePagesObjInfo^Equation Native __1325203681FpSp`pSp`Ole cTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A   MRw2t FMathType 5.0 Equation MathType EFEquation.DSMT49qDSMT5WinAllBasicCodePagesCompObjdiObjInfofEquation Native g_1325203708FpĴp`pĴp`Times New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  kg"m 2 FMathType 5.0 Equation MathType EFEquation.DSMT49qDSMT5WinAllBasicCodePagesOle kCompObjliObjInfonEquation Native oTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  kg"m 2 FMathType 5.0 Equation MathType EFEquation.DSMT49qDSMT5WinAllBasicCodePages_1325203751LF`p``p`Ole sCompObjtiObjInfovEquation Native w_1133241496F`p``p`Ole {CompObj |iTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  kg"m 2 FMathType 4.0 Equation MathType EFEquation.DSMT49q r=mgL 1 -ObjInfo~Ole10Native Equation Native _1214775907 FP:p`P:p`mgL 2 =Ia=mL 12 +mL 22 ()a,2 4r  6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  t==mgL 1 "-mgL 2 ==Ia==mL 12 ++mL 22 ()a, FMathType 5.0 Equation MathType EFEquation.DSMT49qOle CompObj iObjInfoOle10Native a=gL 1 -L 2 ()L 12 +L 22 =9.8()0.20-0.80()0.80 2 +0.20 2 =-8.65rad/s 2= 2 A-A…DSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  a== gL 1 "-L 2 ()L 12 ++L 22Equation Native _1133241258+FPp`@p`Ole CompObji == 9.8m/s 2 ()0.20 m"-0.80 m()(0.20 m) 2 ++(0.80 m) 2 =="-8.65rad/s 2 FMathType 4.0 Equation MathType EFEquation.DSMT49q 6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  a  1 ObjInfoEquation Native _1133241337F@fp`@fp`Ole ==|a|L 1 ==8.65 rad/s 2 ()0.20 m()==1.7 m/s. FMathType 4.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoOle10Native$Equation Native    2a  1 ab=||a|L 1 =8.65 rad/s 2 ()0.80 m()=6.9 m/s 2 2a  2 ab=|a|L 2                 =8.65 rad/s 2 ()0.20 m()                =1.7 m/s. 6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  a  2 ==|a|L 2 ==8.65 rad/s 2 ()0.80 m()==6.9 m/s 2 . FMathType Equation Equation Equation9q a=tI=FrI=0.5t+0.3t 2 ()0.1_1106324313N F@p`@p`Ole CompObj\ObjInfo Ole10Native_1322420947#F0!p`0!p`Ole CompObj"%i0()1.010 -3 =50t+30t 2 FMathType 5.0 Equation MathType EFEquation.DSMT49q@ w=adt=ObjInfoOle10Native$&DEquation Native _1032079826) F0p`0p`25t 2 +10t 3 () 03  30=5.010 2 rad/s.aqlDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  w==1adt==25t 2 ++10t 3 () 03 +"  03 ==5.010 2 rad/s.Ole CompObj(*\ObjInfo+Ole10Native$ FMathType Equation Equation Equation9q  v= 2ad  = 2(2mg)dM+2m  = 4(50)(980)(50)400+2(50)  =1.410 2  cm/s.8  FMathType 5.0 Equation MathType EFEquation.DSMT49q%£DDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A _11324404257.F Mp` Mp`Ole CompObj-/iObjInfo0Equation Native _13224209933F p` p`Ole CompObj25i P==tw!t== 74600 W188.5 rad/s==396 N"m FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfoOle10Native46 Equation Native  _11324404999Fp`p`                    ! " # & ) * , - . / 0 1 4 7 8 9 ; < = > ? @ A B C F I J K N Q T W Z ] ` c f i j m p s v w z }  K=12Iw 2 =1213mL 2 ()w 2 =16mL 2 w 2 .lDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  K== 12Iw 2 == 12 13mL 2 ()w 2 == 16mL 2 w 2 == 16(0.42 kg)(0.75 m) 2 (4.0 rad/s) 2 ==0.63 J.Ole  CompObj8; iObjInfo Ole10Native:<  FMathType 5.0 Equation MathType EFEquation.DSMT49q h=Kmg=mL 2 w 2 6mg=L 2 w 2 6g.cEquation Native  _1220205952@?Fp`p`Ole $ CompObj>A% i%§DDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  h== Kmg== mL 2 w 2 6mg== L 2 w 2 6g== (0.75 m) 2 (4.0 rad/s) 2 6(9.8 m/s 2 )==0.153 mH"0.15 m. FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfo' Ole10Native@B( Equation Native + _1322352091EF4p`4p` I=(32)(1.2) 2 =46.1 kgm 2y‚|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I==(32kg)(1.2 m) 2 ==46.1 kg"m 2 FMathType 5.0 Equation MathType EFEquation.DSMT49q W=DK=0-12Iw 2 =-12(46.1)(29.3)Ole 2 CompObjDG3 iObjInfo5 Ole10NativeFH6 2:$DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  W==DK==0"- 12Iw 2 =="- 12(46.Equation Native : V_1106324369K Fp`p`Ole D CompObjJLE \1kg"m 2 )(29.3 rad/s) 2 =="-1.9810 4  J FMathType Equation Equation Equation9qObjInfoMG Ole10NativeH _1106324523ISP Fp`p`Ole L  |P|=|W|Dt=19.810 3 15=1.3210 3  W.ǟϟϟϟ FMathType Equation Equation Equation9qCompObjOQM \ObjInfoRO Ole10NativeP D_1106324524U F_p`_p`@ l!  FMathType Equation Equation Equation9q@ l/2 Ole R CompObjTVS \ObjInfoWU Ole10NativeV D_1106324525eZ Fp`p`Ole X CompObjY[Y \ObjInfo\[  FMathType Equation Equation Equation9ql  12 mgl,Fce FMathType Equation Equation Ole10Native\ p_1106324526_ Fp`p`Ole ^ CompObj^`_ \Equation9ql  12 Iw 2 ,ll44 FMathType Equation Equation Equation9q    lVW !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUXZY[\^]_`bacdfeghjikmpnoqrtsuvwxyz{|}~ObjInfoaa Ole10Nativeb p_1106324527]gd Fp`p`Ole d CompObjcee \ObjInfofg Ole10Nativeh _1106324528i Fзp`зp` 12mgl=12Iw 2 w= mglI  .<<<< FMathType Equation Equation Equation9q@ l/' Ole k CompObjhjl \ObjInfokn Ole10Nativeo D FMathType Equation Equation Equation9q v=wl= mgl 3 I  ._1106324530bvn F(p`(p`Ole q CompObjmor \ObjInfopt Ole10Nativeu _1106324531s Frp`p`Ole x CompObjrty \ FMathType Equation Equation Equation9q` I= 13 ml 2 FMathType Equation Equation ObjInfou{ Ole10Native| d_1106324532qx Fp`p`Ole ~ CompObjwy \ObjInfoz Ole10Native _1132440606G}Fp`p` Equation9q v= 3gl  = 39.8 m/s 2 ()1.00 m()  = 5.42 m/s. # # # FMathType 5.0 Equation MathTyOle  CompObj|~ iObjInfo Equation Native  pe EFEquation.DSMT49q%µ\DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I==I com ++Mh 2 == 12MR 2 ++Mh 2 == 1220 kg()0.10 m() 2 ++20 kg()0.50 m() 2 ==0.15 kg"m 2 . FMathType Equation Equation _1106324534 FYq`Yq`Ole  CompObj \ObjInfo Equation9q` Mgh= 12 Iw 2 FMathType 5.0 Equation MathType EFEquation.DSMT49qOle10Native d_1214775985FYq`Yq`Ole  CompObj iObjInfo Ole10Native Equation Native  ]_1106324756l F;q`;q` w= 2MghI  = 2(20)(9.8)(0.050)0.15  =11 rad/s.:::AADSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  w==  2MghI  ==  2(20kg)(9.8m/s 2 )(0.050 m)0.15kg"m 2 ==11 rad/s. FMathType Equation Equation rSI}un'n=@C)xڥUKLQM)NiYjӡkҤn%nLJ6tJ"ƕ7Y`Ѹvaܘ2r%1HMgXo}>C@P WRAedrkf\#-.D@8)!6 ŌJT#'c;**Gr=-z/e4I!(OiVҐ>9JؤB@3x@W)fh_>6ソ8|Q-wB74Ax\؉ S[}tf<ຑZÉLj~deAJ$ 8kW)Tө6^8k Kb46wh 2ԏK#a=8"1^ŨӋwHw|4?h$X%B"XX ;bF!%`S ],1 #L33MxfNam ZC=_3hEj,205år ~+dm|=}(1F8  b5BKqg_ʹ[ e Z䏟w*"DdThP  S A? 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FMathType 5.0 Equation MathType EFEquation.DSMT49q$DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  a== 2Rqt ObjInfos Equation Native t _1322352288FtAq`tAq`Ole } 2 == 2(0.024 m)(0.130 rad)(0.0910 s) 2 ==0.754 m/s 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObj~ iObjInfo Ole10Native Equation Native    T 1 =m 1 g-a 1 ()=Mg-2Rqt 2 ().$DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  T 1 ==m 1 g"-a 1 ()==Mg"- 2Rqt 2 ()==(6.20 kg)9.80 m/s 2 "- 2(0.024 m)(0.130 rad)(0.0910 s) 2 ()==56.1 N. FMathType 5.0 Equation MathType EFEquation.DSMT49q T 2 =T 1 -IaR=Mg-2Rqt 2 ()_1322352478RF/Fq`xHq`Ole  CompObj iObjInfo Ole10Native Equation Native  j_13224213151FJq`Jq`Ole  -2IqRt 2?;?;?;?;N$DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  T 2 ==T 1 "- IaR==56.1 N "- (7.4010 "-4 kg"m 2 )(31.4 rad/s 2 )0.024 m==55.1 N. FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObj iObjInfo Equation Native  _1322421334FJq`ZMq`$DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  0==w 0 ++at FMathType 5.0 Equation MathType EFEquation.DSMT49qOle  CompObj iObjInfo Ole10Native  0=w 0 +ata=-w 0 t{z{z?$DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/ED/APG_APA AE%B_AC_AE*_FAGAHA*_D_E_E_A  a=="- w 0Equation Native  _1106325374  FOq`Oq`Ole  CompObj \ t=="- 39.0rev/s32.0s=="-1.22rev/s 2 =="-7.66rad/s 2 FMathType Equation Equation Equation9qObjInfo  Ole10Native d_1106325375  FRq`Rq`Ole  ` Ml 2 /12 FMathType Equation Equation Equation9q@ l  ()CompObj   \ObjInfo Ole10Native D_1106325376 FVq`Vq` FMathType Equation Equation Equation9q` ml/2() 2 ,Ole  CompObj \ObjInfo Ole10Native d_1106325378 F[q`[q`Ole  CompObj \ObjInfo  FMathType Equation Equation Equation9q@ I=Ml 2 12+2ml 2 4=6.40kg()1.20m() 2 12+1.06kg()1.20m() 2 2Ole10Native D_1106325379 Fm`q`m`q`Ole  CompObj \ FMathType Equation Equation Equation9q@  ObjInfo Ole10Native D_1106325380  Fbq`bq`Ole  CompObj! \ObjInfo" Ole10Native _1106325381-% Fgq`gq` FMathType Equation Equation Equation9q t=1.53kgm 2 ()-7.66rad/s 2 ()=-11.7Nm. 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FMathType 5.0 Equation MathTy_1109299028!lZFPNq`PNq`Ole n CompObjY[o iObjInfo\q pe EFEquation.DSMT49qwXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  q A ==q BEquation Native r 2_1109299229_FPq`Pq`Ole w CompObj^`x i FMathType 5.0 Equation MathType EFEquation.DSMT49qwjtXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  w A t== 12a ObjInfoaz Equation Native { _1109299873]gdF@zq`@zq`Ole   B t 2   !  t== 2w A a B == 2(9.5 rad/s)(2.2 rad/s 2 )==8.6 s. FMathType 5.0 Equation MathTyCompObjce iObjInfof Equation Native  _1109300213iF0Ĕq`0Ĕq`pe EFEquation.DSMT49qw,XDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  Dq==q A "-q B ==w A t"- 12a B t 2 ==9.5t"-1.1t 2 . 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FMathType 5.0 Equation MathType EFEquation.DSMT49qwlXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*ObjInfou Equation Native  8_1109301174q0xF q` aq`Ole  _HHAHAE_E_E  t 0 ==8.63 s FMathType 5.0 Equation MathType EFEquation.DSMT49qwXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagCompObjwy iObjInfoz Equation Native  _1214773105N}F Ҡq` Ҡq`essө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  t 0 ==0 FMathType 5.0 Equation MathType EFEquation.DSMT49qADSMT5WinAllBasicCodePagesOle  CompObj|~ iObjInfo Equation Native  2Times New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  t==Ia==rF==rmg. FMathType 5.0 Equation MathType EFEquation.DSMT49q_1214773383Fq`q`Ole  CompObj iObjInfo Equation Native  1_1214773377Fq`q`Ole  CompObj iA DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  I==15kg"m 2 FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfo Equation Native  __1214773697Fקq`קq`Ole  AC DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  a== rFI== rmgI== (0.050 m)(70kg)(9.8m/s 2 )15kg"m 2 ==2.3 rad/s 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49qAžDSMT5WinAllBasicCodePagesCompObj iObjInfo Equation Native  _1214773863Fq`q`Times New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  t net ==Pr i F ii " ==(0.050 m)(70kg)(9.8m/s 2 )"-(0.10 m)(14kg)(9.8m/s 2 )==20.58N"m FMathType 5.0 Equation MathType EFEquation.DSMT49qADSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT ExtOle  CompObj iObjInfo Equation Native  ra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  a== t net I== 20.58N"m15kg"m 2 H"1.4 rad/s 2 . FMathType Equation Equation _1106325926v Fq`sq`Ole  CompObj \ObjInfo Equation9q a=5.0)rads2.0s=2.5)rads 2 . . FF FMathType Equation Equation Ole10Native _1106325927 Fsq`q`Ole  CompObj \               ! $ ' ( ) + , - . / 0 1 2 5 8 9 : ; = > ? @ A B C D E F G J M P S T W Z [ \ ^ _ ` a b c d e f i l m p s v y |  Equation9q` w=Dq/Dt. FMathType 5.0 Equation MathType EFEquation.DSMT49qObjInfo Ole10Native d_1132442058:Fq`q`Ole  CompObj iObjInfo Equation Native  1_1220206422jFq`Zq`%Tt,DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  q 1 == 12at 2  t==20 ==500rad,w==at t==20 ==50 rads. FMathType 5.0 Equation MathType EFEquation.DSMT49q q 2 =q 1 +wDt=500+50()20()=1500rad. Ole  CompObj iObjInfo Ole10Native y|DSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  q 2 ==q 1 ++wDt==500 rad++50 rad/s()20s()==1.Equation Native  _1109035104F˿q`˿q`Ole " CompObj# i510 3 rad. FMathType 5.0 Equation MathType EFEquation.DSMT49q a=-33.33 rev/min0.50min=-66.7 rev/miObjInfo% Ole10Native& Equation Native * _1220206652Fq`q`n 2 .|XDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT ExtraWinAllCodePagessө!/ED/APG_APA@AE%B_AC_AE*_FAGAHA*_D_E_E_A  a=="- 33.33 rev/min0.50min=="-66.7 rev/min 2 H""-67 rev/min 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49q q=w 0 t+12atOle 3 CompObj4 iObjInfo6 Ole10Native7  2 =(33.33)(0.50)+12(-66.7)(0.50) 2 =8.3 rev.y«|DSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclidMath1Euclid Math OneMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  q==Equation Native < _1106325530B Fq`q`Ole H CompObjI \w 0 t++ 12at 2 ==(33.33 rev/min)(0.50 min)++ 12("-66.7rev/min 2 )(0.50 min) 2 ==8.3 rev. FMathType Equation Equation Equation9q`  12 Iw 2 FMathType Equation Equation ObjInfoK Ole10NativeL d_1106325532 FAq`Aq`Ole N CompObjO \ObjInfoQ Ole10NativeR _1214776215Fq`q`Equation9q 0=-MgL/2+12Iw 2 w= MgL/I  .!!!!! FMathType 5.0 Equation MathType EFEquation.DSMT49qOle U CompObjV iObjInfoX Ole10NativeY  w= MgLI  = MgL4ML 2 /9  = 9g4L  .AWDSMT5WinAllBasicCodePagesTimes New RomanSymbolMT Extra!/ED/APG_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A Equation Native ] s_1106325563 Fmq`mq`Ole g CompObjh \ w==  MgLI  ==  MgL4ML 2 /9  ==  9g4L  ==  9(9.800 m/s 2 )4(0.600 m)  ==6.06 rad/s. FMathType Equation Equation Equation9q I C =12MR 2 andI H =Mr 2 .an W FMathType Equation Equation ObjInfoj Ole10Nativek _1106325564 Fq`q`Ole n CompObjo \ObjInfoq Ole10Nativer d_1106325565 F(q`(q`Equation9q` R 2 /2=R H2y FMathType Equation Equation Equation9qOle t CompObju \ObjInfow Ole10Nativex d` R H =R/ 2  FMathType Equation Equation Equation9q` I=Mk 2_1106325566 Fq`q`Ole z CompObj{ \ObjInfo} Ole10Native~ d_1106325567 FTq`q`Ole  CompObj \  FMathType Equation Equation Equation9ql k= I/M ll@KK FMathType 5.0 Equation MathTyObjInfo Ole10Native p_1132442303Fpq`pq`Ole  CompObj iObjInfo Equation Native  _1106325708 Fpq`pq`pe EFEquation.DSMT49q%tDSMT5WinAllBasicCodePagesTimes New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A   w 12 2aOle  CompObj \ObjInfo Ole10Native d FMathType Equation Equation Equation9q` h= L2 sin40 FMathType Equation Equation Equation9q_1106325709VX F`q``q`Ole  CompObj \ObjInfo `  12 Iw 2> FMathType Equation Equation Equation9q I= 112 ML 2 +MOle10Native d_1106325710 F`q``q`Ole  CompObj \ObjInfo Ole10Native _1132441903FPgq`Pgq`Ole  (L/2) 2 = 13 ML 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49q%[lDSMT5WinAllBasicCodePagesCompObj iObjInfo Equation Native  w_1106324910 F@q`@q`Times New RomanSymbolEuclid Extra!/'_!!/G_APA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  Mg L2sin40== 12 13ML 2 ()w 2 !w==  3gsin40L  ==3.1 rad/s. FMathType Equation Equation Equation9q I=mR 2 =NmR 2 Ole  CompObj \ObjInfo Ole10Native _1106324911F@q`@q`Ole  CompObj iObjInfo  FMathType 4.0 Equation MathType EFEquation.DSMT49q W=12Iw 2 =12NmR 2 w 2 .```X@6M6GxDSMT4WinAllBasicCodePagesOle10Native Equation Native  t_1106324912  F0q`0q`Ole  Times New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  W== 12Iw 2 == 12NmR 2 w 2 FMathType Equation Equation Equation9qCompObj \ObjInfo Ole10Native $_1106325226  F00r`00r`  122160()50()38() 2 0.052() 2 W122160()100()38() 2 0.052() 2 210 5 JW410 5 J. FMathType Equation Equation Equation9q` a 1 =a 2 =Ra>;>;>;Ole  CompObj  \ObjInfo  Ole10Native d_1106325227B F zr` zr`Ole  CompObj  \ObjInfo  FMathType Equation Equation Equation9q T 1 -m 1 g=m 1 a 1 m 2 g-T 2 =m 2 a 2 T 2 R-T 1 R=IaOle10Native _1106325228 F5r`5r`Ole  CompObj \ FMathType Equation Equation Equation9qh I= 12 MR 2{{{{{{{hOhOObjInfo Ole10Native l_1106325229 F r` r`Ole  CompObj \ObjInfo Ole10Native _1325204091Far`ar` FMathType Equation Equation Equation9q m 2 g-m 1 g=m 1 +m 2 +12M()a FMathType 5.0 Equation MathTy    #" `2Iuan$Oۦ_u޿%`!uan$Oۦ_u޿ @3@rOxڵVMlU&8v0q$q쵒nmJ"(G+ BH(W"*R\P HADD }vcc77[n1AngT*fXV񄖈udȀ#%?,kVdrώ.<Z0'khDŃj3y]]tuۮbTMݭbGv禺_@/I}$3Y(<=; Fx==~r(,}fD@Ȩs{L]PU_FC^cy}8.`_rMt<jPGb:mPAxݰz1:UOS lcs\+L|;cqxmdчu/}&YrSE\Qܺ]߰[.4ug\Eh4rj@`rhXF o JG; +pirl<[UUqꏮdž?n]PŶEǏfNXߵ<:|ʇldu:XLoeѲƺQZox<0{Pojۿ;h_џQZ;iu[ju71hb CS3C3g'ù|:j_wh`tBK@z15.}G_ܟL}I(ei1]ޟ}hFx~ tlL{ư͌OuB\{-q+ h)GnvQp0=hy03ԏWM1_P;2'`#CXū>جCUK2gN2l&'aI-dVĔ`8sdrd& )nmCIv+樭M@~Lџ܁6\ftŜ\hd8/7z ;7 //0 1Ddl\ % c $A%? ?#" `2\)%Sq=``!\)%Sq=` xxڵTkA3IMB@,ڀǀ4K$VȇIV6H!zxK8*4NQnQ),QYFr|",̦z]GX( ~ܖʙO>=x=` i}тΜ̉ږfkptO nba`(U qDdJ  C A? 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FMathType Equation Equation Equation9q t=rf=0.780m()ObjInfoaNOle10NativeO_1106325995ld Fr`r`Ole S2.3010 -2 N()=1.7910 -2 Nm.ǀǀǀǀ FMathType Equation Equation Equation9q@ l*> CompObjceT\ObjInfofVOle10NativeWD_1106325996i FЇr`Їr` FMathType Equation Equation Equation9q I=I s +I l =112m s L s2 +13m l L l2 =0.019kOle YCompObjhjZ\ObjInfok\Ole10Native]gm 2 . FMathType Equation Equation Equation9q I=I s +I l =112m s L s2 +m s _1106325997gqn Fr`r`Ole bCompObjmoc\ObjInfopeOle10Nativef_1106325998s FBr`r`Ole kCompObjrtl\h 2 +112m l L l2???? FMathType Equation Equation Equation9q@ 90<ObjInfounOle10NativeoD_1133239879?xF$r`$r`Ole q FMathType 4.0 Equation MathType EFEquation.DSMT49q» 6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_ECompObjwyriObjInfoztEquation Native u_1133240066}Fnr`r`_A  v 1 ==r A w A ==(15 cm)(10 rad/s)==1.510 2  cm/s FMathType 4.0 Equation MathType EFEquation.DSMT49qOle }CompObj|~~iObjInfoEquation Native i 6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  r B w B ==r A w A    ! w B == r A w A r B == 15 cm10 cm()(10 rad/s)==15 rad/s FMathType 4.0 Equation MathType EFEquation.DSMT49q_1133240257{F)r`)r`Ole CompObjiObjInfoEquation Native _1133240367Fs`s`Ole CompObji 6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  B FMathType 4.0 Equation MathType EFEquation.DSMT49qObjInfoEquation Native *_1133240437FUs`Us`Ole  6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  w B ==15 rad/s FMathType 4.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoEquation Native _1220206962F7 s`7 s`— 6M5GDSMT4WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  v 2 ==r B w B ==(5 cm)(15 rad/s)==75 cm/sOle CompObjiObjInfoEquation Native  FMathType 5.0 Equation MathType EFEquation.DSMT49qyyDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HA@AHA*_D_E_E_A  r C w C ==r B w B    ! w C == r B w B r C == 5 cm25 cm()(15 rad/s)==3.0 rad/s_1106325179G F s` s`Ole CompObj\ObjInfo FMathType Equation Equation Equation9q I=m i r i2  =2M()L 2 +2M()L 2 +M2L() 2????Ole10Native_1106325181 Fs`p<s`Ole CompObj\ FMathType Equation Equation Equation9q@ 90< FMathType Equation Equation Equation9qObjInfoOle10NativeD_1106325182 Fps`ps`Ole CompObj\ObjInfoOle10Native_1106325183 Fps``hs` K=12Iw 2 =3.3J.{{{{{{{p FMathType Equation Equation Equation9qOle CompObj\ObjInfoOle10Native I=m i r i2  =2M()r 2 +2M()r 2 +M2L() 2P30 FMathType Equation Equation Equation9q_1106325184 F`hs``s`Ole CompObj\ObjInfoOle10NativeD_1106325185 F`Js`P s`Ole CompObj\@ 90< FMathType Equation Equation Equation9q K=12Iw 2 =2.9J.5"&ObjInfoOle10Native_1109319765FP#s`P#s`Ole  FMathType 5.0 Equation MathType EFEquation.DSMT49qwlXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*CompObjiObjInfoEquation Native _1109320375F@'s`@'s`_HHAHAE_E_E  I== 112ML 2 ++Mh 2 == 112(3.0 kg)(4.0 m) 2 ++(3.0 kg)(1.0 m) 2 ==7.0 kg"m 2 .Ole CompObjiObjInfoEquation Native _ FMathType 5.0 Equation MathType EFEquation.DSMT49qwCtXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  K rot == 12Iw 2   ! w=  2K rot I  ==  2(20 J)7 kg"m 2 ==2.4 rad/s FMathType 5.0 Equation MathTy_1109320446F0{,s`0{,s`Ole CompObjiObjInfo   !$'()*-012369<?@ADGHIJKLMPSTVWXYZ[]^_`abdefghijklmnppe EFEquation.DSMT49qw|XDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  v B ==wr AB ==(2.4 rad/s)(3Equation Native _1109321101F0]1s`0]1s`Ole  CompObj i.00 m)==7.2 m/s FMathType 5.0 Equation MathType EFEquation.DSMT49qw8lXDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagObjInfo Equation Native T_1132442363F 6s` 6s`Ole essө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  Dy==d AC (1"-cosq) FMathType 5.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfoEquation Native {_1109321273Fb8s`b8s`%_tDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  DU==mgDy==mgd AC (1"-cosq)!20 J==(3.0 kg)(9.8 m/s 2 )(1.0 m)(1"-cosq) FMathType 5.0 Equation MathType EFEquation.DSMT49qw XDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT ExtraWinAllCodePagOle "CompObj#iObjInfo%Equation Native &,essө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  cosq==0.32 FMathType 5.0 Equation MathType EFEquation.DSMT49qw|XDSMT5WinAllBasicCodePages_1109321289DF:s`:s`Ole +CompObj,iObjInfo.Equation Native /_1106325734 FD=s`?s`Ole 4CompObj5\Times New RomanSymbolCourier NewMT ExtraWinAllCodePagessө!/ED/APG_APAPAE%B_AC_AE*_HHAHAE_E_E  qH"71 FMathType Equation Equation Equation9qObjInfo7Ole10Native8d_1106325735 FAs`As`Ole :` r= 2  L=0.71m FMathType Equation Equation Equation9q I=m i r i2  =2mL 2 +m 2  L()CompObj;\ObjInfo=Ole10Native>_1133201540FpDs`pDs` 2 =0.20kgm 2 . FMathType 5.0 Equation MathType EFEquation.DSMT49q-dDSMT5WinAllBasicCodePagesTimes New RomanSymbolCourier NewMT Extra!/'_!!/G_Ole BCompObjCiObjInfoEEquation Native FAPA@AE%B_AC_AE*_FA@AHA*_D_E_E_A  K 0 ++U 0 ==K++U!0++4m()gh 0 ==K++0. 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