ࡱ> #` I_bjbjmm 4l#%DDD8DT"E|X_ EE"EEEEEEG^I^I^I^I^I^I^$x`hbm^HEEHHm^EE_uLuLuLHEEG^uLHG^uLuL\]EE 𾕂WXDI ?]G^(_0X_S]ncJ$nc(]nc]PEvBFTuLFDF;EEEm^m^LXEEEX_HHHH&++ Chapter 7 and 8 notes Teachers Copy Chapter 7 Rotational Motion and The Law of Gravitation Rotational Motion motion of a body that spins about an axis Axis of rotation line about which rotation occurs Perpendicular to motion Through center of motion Linear equations will not work b/c direction of rotational motion is constantly changing. Analyzing circular motion describe through angles of motion all points on rotating body move through same angle set up a fixed reference line r = distance from center  = angle from reference line s = arc length  distance moved along the circumference of circle. Measuring Angles Use radians  angle whose arc length (s) is equal to the radius (r)  = s r Pure number therefore, use (rad) as label 1 revolution (s) = 2r One complete revolution = 360o  = s = 2r = 2  rad r r To convert angle in degrees to radians  (rad) =   (deg.) 180o Practice converting the following angles to radians. 25o 35o 128o 270o Angular Displacement  angle through which a point, line, or body is rotated in a specific direction and about a specific axis. " (rad) = "s r (+) "s = counterclockwise rotation (-) "s = clockwise rotation Practice Problems While riding a carousel that is rotating clockwise, a child travels through an arc length of 11.5 m. If the childs angular displacement is 165o, what is the radius of the carousel? Earth has an equatorial radius of approximately 6380 km and rotates 360o every 24 h. What is the angular displacement (in degrees) or a person standing at the equator for 1.0 h? Angular Speed rate that a body rotates around an axis  = greek letter omega avg = " (rad) "t SI = rad / s 1 revolution = 2 rad Angular Acceleration  rate of change of   = Greek letter alpha avg = 2  1 = " or (a/d) ! tangential acceleration (m/s2) / distance (m) t2  t1 "t SI = rad/s2 All points on a rigid body have the same  and  Pg 251 has a table that compares rotational and linear kinematic equations Practice Problems A car s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 s the tires angular speed is 28.0 rad/s. What is the tires average angular acceleration during the 3.5 s time interval? A top that is spinning at 15 rev/s spinning for 55 s before coming to a stop. What is the average angular acceleration of the top while it is slowing? Newtons Law of Gravitation Gravitational Force attractive force between two objects Increased mass = increased Fg Increased distance between = decreased Fg Fg = G (m1)(m2) r2 G =Constant of Universal Gravitation = 6.67 x 10-11 N m2 kg2 Calculate Fg of an object on the surface of a spherical object Fg = G (ME)(m2) (RE)2 *Remember Fg = (m) (g) Practice Problems Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the Fg is 8.92 x 10-11 N. Find the Fg exerted on the moon (mass = 7.36 x 1022 kg) by Earth (mass = 5.98 x 1024 kg) when the distance between them is 3.84 x 108 m. Chapter 8 Rotational Equilibrium and Dynamics Torque the ability of a force to rotate an object around some axis Net torque produces rotation Occurs around an axis of rotation usually a hinge. Imaginary line passing through a hinge Torque depends on force and lever arm Ease of rotation depends on: How much force is applied Where the force is applied Farther from the axis of rotation the easier the rotation More torque produced Lever arm perpendicular distance from axis of rotation to a line drawn along the direction of the force Distance of lever arm = d Shorter lever arm = smaller torque Torque depends on the angle between force and the lever arm Changing the angle will change the ease that the object will move with Torque = (() = Greek letter Tau SI unit = N m Torque is (+) or (-) depending on direction of the rotation (+) When rotation is counterclockwise (-) When rotation is clockwise When more than 1 force causes rotation then net torque is the sum of all torques involved. Section 8-2 Center of Mass Center of Mass point at which all of the mass of the body can be considered to be concentrated when analyzing transitional motion. Regular shaped objects (i.e. sphere, cube) center of mass is at the geometric center of the object. Different for oddly shaped objects Average position of an objects mass Center gravity an average position at which the gravitational force of the object acts. In this book Center of Mass and Center of Gravity are equivalent. Moment of Inertia Mini Lab: Moment of inertia of a rod Pg 285 Calculate I for rod in each position Why is it easier to rotate the rod around some axis and not others? The moment of inertia changes as hand position changes. Moment of Inertia measure of the resistance of an object to changes in rotational motion (I) Indicates how much an objects rotation will change with a force Depends on: Objects mass Distribution of mass around the axis of rotation Farther the mass is from the axis of rotation More difficult to rotate Greater moment of inertia Closer mass is to axis of rotation Less difficult to rotate Smaller moment of inertia Calculate Moment of Inertia When net torque acts on an object, the resulting change in rotational motion depends on the objects moment of inertia Table 8-1 helps you calculate the moment of inertia for common shapes SI units Kg m2 Only the distance of a mass from the axis of rotation is important in determining the moment of inertia for a shape Rotational equilibrium If 2 forces acting on an object are equal in magnitude and opposite in direction, the object will rotate in place. Translational equilibrium net force on an object is equal to zero Rotational equilibrium net torque on an object is equal to zero Rotational Dynamics Newtons 2nd law for Rotating Objects (net = I I = moment of inertia  = angular acceleration (+) or (-) depending on direction of ( Translational vs. Rotational F = ma (net = I Practice Problems A student tosses a dart using only the rotation of her forearm to accelerate the dart. The forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart have a combined moment of inertia of 0.075 kg m2 about the axis, and the length of the forearm is 0.26 m. if the dart has a tangential acceleration of 45 m/s2 just before its is released, what is the net torque on the arm and dart? Momentum Angular momentum (L)- product of a rotating objects moment of inertia and angular speed around the same axis L = I I = moment of inertia  = angular speed Kg " m2/s ! derived from (Kg " m2) (rad/sec) (Rad have no value) L may be conserved Law of conservation of L  When net = 0 and "L = 0 A skater spinning with arms out spins slower than when he/she brings arms in More mass is brought closer to the body therefore decreasing  I therefore, increasing   to compensate for decreased  I ,  L is conserved Practice Problem A 65 kg student is spinning on a merry-go-round that has a mass of 5.25 x 102 kg and a radiu +-du L  , 0 l n p x z | 걢|tߖhߖ|||ߖߖߖhvwh}nCJH*aJhvwCJaJhvwh}n>*CJaJhvwh}n>*CJaJo(hvwh}nCJaJo(hvwhC^56>*CJaJhvwhC^CJaJhvwh'cmCJaJo(hvwh'cm56>*CJaJhACJaJhvwh}nCJaJhvwh'cmCJaJhvwhvCJaJ(,-d a {  L  2 p  & Fgd}n^gd}n ^`gdvw & FgdC^ & Fgd'cm & Fgd'cm & Fgd'cm#_H_p & b x z   & Fgd}ngd}n@ ^@ gd}n & Fgd}n & FgdC^^gd}np^pgd}n ^`gd}n & ( * < > D H N P ` t x @06 ÷}uh_ CJaJhvwh_ CJH*aJhvwCJaJhvwh}n56>*CJaJhvwh}nCJH*aJhvwh_ CJaJhvwh_ >*CJaJhvwh}n>*CJaJhAhA>*CJaJhAh}n>*CJaJo(hvwh}nCJaJo(hvwh}nCJaJ)6 2. & Fgd_  & Fgd'cm & Fgd_ & F]gd_ gd_  & Fgd_  & Fgd}ngd}n & Fgd}n.06<>LN8:<DN꼮꼁vg[PhvwhrCJaJhvwh[fCJaJo(hvwh[f56>*CJaJhvwh^CJaJhvwh[fCJaJhvwCJaJhvwh_ >*CJaJo(hvwh_ >*CJaJhvwh_ CJH*aJo(hvwh_ CJaJo(hvwh}nCJaJhvwh_ 56>*CJaJh_ CJaJhvwh_ CJaJhvwhACJaJ.N<j^vnpgdr & Fgdr & Fgdr^gd^ & Fgd[f8^8gdvw & Fgdvw & Fgd[f & Fgd'cm8^8gd^gdvw & Fgd_ Nlrxz|6:<DF\^rtvlnpPQRTpϸ|qf^qfhrCJaJhvwhrCJaJhvwh^CJaJh[fCJaJhvwh[fCJH*aJhvwh[fCJaJo(hvwCJaJh^h^CJH*aJh^CJaJh^>*CJaJhvwh[fCJH*aJo(hvwh[f>*CJaJo(hvwh[f>*CJaJhvwh[fCJH*aJhvwh[fCJaJ#MNOPQRSpK@ ^@ gdr^gd[f@ ^@ gd[f & Fgd[f & Fgd[f & Fgd'cm & Fgdrgdrp  @CDIJͫ堔堔Š|hvwhrCJH*aJhvwhr>*CJaJhvwhrCJH*aJhvwhrCJaJhvwh[f>*CJH*aJhvwh[fCJH*aJh^CJaJhvwh[f>*CJaJhvwh[fCJH*aJhvwh[fCJaJhvwh[f56>*CJaJ.()*+,-./01`agdr & Fgdr & Fgdr^gdr~!",-./1_ag =Kq ⷬxqihw=CJaJ hThThvCJaJ jthvwhvCJaJhvwhv56>*CJaJhvwhw=56>*CJaJhvwhvCJaJhvwhw=CJaJh^CJaJhvwhrCJH*aJhvwhrCJH*aJhvwhrCJaJhvwh^CJaJhrCJaJ(a Fc}Qk2B~ & F & F & F8^8 & F & Fgdw= & Fgdw= <=(Kpq  F M t K! & Fgdw= & F & Fgdw= & F8^8^ & F & Fh^hgdw= & Fgdw=gdT ( F t !K!Q!""J#K#}#~## $~$$$$%%#%%%@%A%D%&&&&' ''ĵĩĵĵϚ{od{d{odYhvwhACJaJhvwhJCJaJhvwhJCJH*aJ jthvwhJCJaJ hvwhJ56>*CJH*aJhvwhJ56>*CJaJhvwhvCJH*aJhvwhv56>*CJaJhvwhvCJaJhvwhw=CJaJhvwhv5CJaJhvwhw=5CJaJhvwhv56CJaJ"K!Q!!!!! 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Section 8-4 Simple Machines Machine any device that transmits or modifies force usually by changing the force applied to an object Made of a combination of simple machines Simple Machine change the direction or magnitude of an input force 6 types lever pulley inclined plane wheel and axle wedge screw Simple Machine Hand-out Shows each simple machine and how to calculate the ideal mechanical advantage for each. Characterized by comparing how large the input force is relative to the output force = Mechanical Advantage MA = Output force = Fout Input force = Fin Consider a hammer: Input force is applied to one end of the handle Output force is exerted by the handle in the head of the nail Rotational equilibrium is maintained therefore, input torque must equal output torque. (in = (out Fin din = Fout dout Mechanical advantage may then be calculated by: MA = Fout = din Fin = dout Longer the input lever as compared to the shorter output lever will result in greater MA Machines can alter the force and the distance moved A machine can increase (or decrease) the force acting on an object at the expense (or gain) or the distance moved Small distance = large force Large distance = small force Work done on an object is always constant Efficiency is a measure of how well a machine works Measure of how much input energy is lost compared with how much energy is used to perform work on an object. 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L^`LhH. ^`hH. ^`hH. PLP^P`LhH.h^`5o(.h ^`hH.h pLp^p`LhH.h @ @ ^@ `hH.h ^`hH.h L^`LhH.h ^`hH.h ^`hH.h PLP^P`LhH.  ^`o(hH.8 ^`o(hH.pLp^p`L.@ @ ^@ `.^`.L^`L.^`.^`.PLP^P`L.h^`5o(.h pp^p`hH.h @ L@ ^@ `LhH.h ^`hH.h ^`hH.h L^`LhH.h ^`hH.h PP^P`hH.h  L ^ `LhH.v! N!p#<;_HB=z_$#JL.Rlվ      L.        L.         [fJ'cmvwvTw=C^r_ ^Ar}n@##LW##!$P@P P PP @PPP,@P&PP@P*P,P@UnknownG: Times New Roman5Symbol3& : Arial3Times5& zaTahoma 1hSfrfrf^A^A4d##f `HX ?'cm2Section 8-2 Center of Mass Gary Anderson jayme.fast(       Oh+'0  < H T `lt|Section 8-2 Center of MassGary AndersonNormal jayme.fast7Microsoft Office Word@ @NsWX@Tmpt@NsWX^՜.+,0 hp  St. James Public SchoolsA#' Section 8-2 Center of Mass Title  !"#$%&'()*+,-./012345689:;<=>@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnoprstuvwxz{|}~Root Entry FХWXData 71Table?cWordDocument4lSummaryInformation(qDocumentSummaryInformation8yCompObjq  FMicrosoft Office Word Document MSWordDocWord.Document.89q