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Chapter 8 Rotational Equilibrium and Rotational Dynamics Wrench Demo Torque Torque, t , is the tendency of a force to rotate an object about some axis t Fd F is the force d is the lever arm (or moment arm) Units are Newton-m Direction of Torque Torque is a vector quantity Direction determined by axis of twist Perpendicular to both r and F In 2-d, torque points into or out of paper Clockwise torques point into paper. Defined as negative Counter-clockwise torques point out of paper. Defined as positive Non-perpendicular forces Only the ycomponent, Fsinf, produces a torque Non-perpendicular forces t Fr sin f F is the force r is distance to point where F is applied Φ is the angle between F and r Torque and Equilibrium Forces sum to zero (no linear motion) Fx 0 and Fy 0 Torques sum to zero (no rotation) t 0 Meter Stick Demo Axis of Rotation Torques require point of reference Point of reference can be anywhere Use same point for all torques Pick the point to make problem least difficult Example Given M = 120 kg. Neglect the mass of the beam. a) Find the tension in the cable b) What is the force between the beam and the wall Solution a) Given: M = 120 kg, L = 10 m, x = 7 m Find: T Basic formula t 0 Axis TL Mgx 0 Solve for T = 824 N Solution b) Given: M = 120 kg, L = 10 m, x = 7 m, T = 824 N Find f Basic formula F y 0 T Mg f 0 Solve for f = 353 N f Alternative Solution b) Given: M = 120 kg, L = 10 m, x = 7 m Find f Basic formula t 0 Mg ( L x) fL 0 Solve for f = 353 N f Axis Another Example Given: W=50 N, L=0.35 m, x=0.03 m Find the tension in the muscle W x Basic formula t 0 L Fx WL 0 F = 583 N Center of Gravity Gravitational force acts on all points of an extended object However, it can be considered as one net force acting on one point, the center-of-gravity, X. (m g ) x i i X mi g i i X mi xi i m i i Weighted Average Example Consider the 400-kg beam shown below. Find TR Solution Find TR Basic formula t 0 to solve for TR, choose axis here MgX TR L 0 TR = 1 121 N X=2m L=7m One Last Example Given: x = 1.5 m, L = 5.0 m, wbeam = 300 N, wman = 600 N Find: T Fig 8.12, p.228 Slide 17 x L Solution Given: x , L , wbeam ,wman Find: T First, find Ty Basic formula Fig 8.12, p.228 t 0 Slide 17 wman x wbeam L / 2 Ty L 0 Ty = 330 N Now, find T Ty T sin( 53) T = 413 N x L Baton Demo Moment-of_Inertia Demo Torque and Angular Acceleration When t 0, rigid body experiences angular acceleration Relation between t and a is analagous to relation between F and a F ma, t Ia Moment of Inertia Moment of Inertia This mass analog is called the moment of inertia, I, of the object I mi ri i 2 r is defined relative to rotation axis SI units are kg m2 More About Moment of Inertia I depends on both the mass and its distribution. If an object’s mass is distributed further from the axis of rotation, the moment of inertia will be larger. Demo: Moment of Inertia Olympics Moment of Inertia of a Uniform Ring Divide ring into segments The radius of each segment is R I mi ri MR 2 2 Example What is the moment of inertia of the following point masses arranged in a square? a) about the x-axis? b) about the y-axis? c) about the z-axis? Solution a) Find I about the x-axis? Given: M2=2, M3=3 kg, L=0.6 m Basic formula I mi ri 2 r = 0.6·sin(45) First, find distance to 2-kg masses I M 2r M 2r 2 2 =0.72 kg·m2 Solution b) Find I about the y-axis? r = 0.6·sin(45) Same as before, except you use the 3-kg masses I M 3r M 3r 2 2 =1.08 kg·m2 Solution c) Find I about the z-axis? r = 0.6·sin(45) Use all the masses I M 2r M 2r 2 2 M 3r M 3r 2 2 =1.8 kg·m2 Other Moments of Inertia Example Treat the spindle as a solid cylinder. a) What is the moment of Inertia of the spindle? b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel? c) What is the acceleration of the bucket? d) What is the mass of the bucket? M Solution a) What is the moment of Inertia of the spindle? Given: M = 5 kg, R = 0.6 m Moments of Inertia 2 MR , cylindrica l shell 1 MR 2 , solid cylinder 2 2 MR 2 , solid sphere 5 2 2 MR , spherical shell 3 1 ML2 , rod, about end 3 1 ML2 , rod, about middle 12 M 1 I MR 2 = 0.9 kgm2 2 Solution b) If the tension in the rope is 10 N, what is a? Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m Basic formula t rF t Ia rT Ia Solve for a a=6.67 rad/s2 c) What is the acceleration of the bucket? Given: r=0.6 m, a = 6.67 rad/s Basic formula a ar a=4 m/s2 M Solution d) What is the mass of the bucket? Given: T = 10 N, a = 4 m/s2 Basic formula F ma Ma Mg T T M g a M = 1.72 kg M Example A cylindrical space station of radius R = 12 m and mass M = 3400 kg is designed to rotate around the axis of the cylinder. The space station’s moment of inertia is 0.75 MR2. Retrorockets are fired tangentially to the surface of space station and provide an impulse of 2.9x104 N·s. a) What is the angular velocity of the space station after the rockets have finished firing? b) What is the centripetal acceleration at the edge of the space station? Solution Given: M = 3400, R = 12, I = 0.75 MR2 = 3.67x105 F·t = 2.9x104. a) Find: w Basic formula w t Ia I t Basic formula t rF t t Iw RF t Iw Solve for w. w= 0.948 rad/s b) Find centripetal acceleration Basic formula 2 a w r a w R 2 = 10.8 m/s2 Rotational Kinetic Energy Each point of a rigid body rotates with angular velocity w. 1 1 1 2 2 2 2 KE mi vi mi ri w Iw 2 2 2 Including the linear motion 1 2 1 2 KE mv Iw 2 2 KE due to rotation KE of center-of-mass motion Example What is the kinetic energy of the Earth due to the daily rotation? Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m. Basic formula 2 w T Solid sphere 2 I MR 2 5 Basic formula 1 2 KE Iw 2 First, find w 2 = 7.27 x10-5 rad/s w 24 3600 1 2 2 KE MR w = 2.78 x10-29 5 Example A solid sphere rolls down a hill of height 40 m. What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!) Basic formula 1 2 1 2 mgh mv Iw 2 2 2 2 m cancels I mr 1 2 2 2 2 5 gh v r w 2 5 For solidsphere Basic formula v wr 1 2 2 2 2 2 gh gh v v , v 2 5 7/5 v = 23.7 m/s Suitcase Demo Angular Momentum L Iw L mvr Rigid body Point particle Analogy between L and p Angular Momentum Linear momentum L = Iw p = mv t = L/t F = p/t Conserved if no net outside torques Conserved if no net outside forces Rotating Chair Demo Angular Momentum and Kepler’s 2nd Law For central forces, e.g. gravity, t = 0 and L is conserved. Change in area in t is: 1 A r (v t ) 2 L mrv A 1 L t 2m Example A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially. Solution Known: M, R, m, v0 Find: wF First, find L0 Basic formula L mvr L0 mvR Next, find Itot Solid cylinder I Tot 1 2 mR MR 2 2 1 2 I MR 2 Now, given Itot and L0, find w Particle I MR 2 Basic formula L Iw L0 w I Tot = 2.71 rad/s Example Two twin ice skaters separated by 10 meters skate without friction in a circle by holding onto opposite ends of a rope. They move around a circle once every five seconds. By reeling in the rope, they approach each other until they are separated by 2 meters. a) What is the period of the new motion? b) If each skater had a mass of 75 kg, what is the work done by the skaters in pulling closer? Solution a) Find: Tf Given: R0=5 m, Rf=1 m, T0=5 s Basic formula L Iw 2 w T Point particle I mr 2 First, find expression for initial L (2 skaters) 2 2 L0 2mR0 T0 Next, apply conservation of L 2 2 2 2mR 2mR f T0 Tf 2 Rf T f T0 2 = T0/25 = 0.2 s R0 2 0 Solution b) Find: W Given: m=75 kg, R0=5 m, Rf=1 m, T0=5 s, Tf=0.2 s First, find moments of inertia I 0 2mR , I f 2mR 2 0 2 f Next, find values for w 2 2 w0 , wf T0 Tf Basic formula 1 2 KE Iw 2 Finally, find change in KE 1 1 2 W I f w f I 0w 02 = 7.11x105 J 2 2