1 12. Rolling, Torque, and Angular Momentum

1 112. Rolling, Torque, and Angular MomentumRolling Motion: A motion that is a combination of rotationaland translational motion, a wheel rollingdown the road. Will only consider rolling with out a disk or sphere rolling along a horizontalsurface, the motion can be considered in twoways: of rotational and translationalmotion: Center of mass moves in a translationalmotion. The rest of the body is rotating around thecenter of cmvvcmfifi II. Pure Rotational Motion:2 The whole object is revolving around apoint on the object in contact with thesurface. The point of contact changes with time. Most people find method I simpler gndfiaxis of rotationUse method I to analyze rolling withoutslipping:RRd= 2 When the object makes one completerevolution, the object has moved a distanceequal to the circumference, and each point onthe exterior has touched the ground once.

2 When the object rotates through an angle ,the distance that the center of mass hasmoved is:3s=R vcm=Rd dt=R where is the Angular velocity of one objectrotating about its center of mass. This looks verysimilar to the relationship between angularvelocity and the translational velocity of a pointon a rotating object:v=R vcm is the velocity of the center of mass withrespect to the ground for the rolling motion. v is the velocity of a point on the object withrespect to the axis of velocity of any point on the disk as seenby an observer on the ground is the vector sumof the velocity with respect to the center of massand the velocity of the center of mass withrespect to the ground: r v gnd=r v relcm+r v cm(1)4rel cmvvvvcmcmrel gndfifififiConsider the point on the top of the wheel:Rvvvcmcmrel cmfififivrelcm=+R vcm=R (1): vgnd=R +vcm=vcm+vcm=2vcm5 The point on the top of the wheel has a speed(relative to the ground) that is twice thevelocity of the center of the point in contact with theground.

3 Rvvvcmcmrel cmfififivrelcm= R vcm=R vgnd= R +R =0 The point in contact with the ground has aspeed of zero, momentarily at your car is traveling down the highway at70 mph, the tops of your wheels are going 140mph while the bottoms of the wheels aregoing 0 a disk rolling down a ramp withoutslipping:6h RAssuming the disk is initially at rest: What makes the disk start rolling? What is the translational speed of its center ofmass when it reaches the bottom of the ramp? What is its Angular velocity when it reachesthe bottom of the ramp?fNmgHWe need to find the torque that makes thedisk start rolling. Consider the rotationaround the center of mass: The force of gravity acts at the center ofmass and hence produces no torque. The normal force has zero level arm andhence produces no torque. The force of friction provides the torque!7 If the object is rolling without slipping, thefriction force is static friction.

4 If the ramp is frictionless, the disk willslide down without velocity and Angular velocity at thebottom of the ramp can be calculated usingenergy conservation. The kinetic energy canbe written as a sum of translational androtational kinetic energy:Ktot=Ktrancm+Krotreltocm=12mvcm2+ 12 Icm 2where is the Angular speed of the rotationrelative to the center of mass and Icm is themoment of inertia around an axis passingthrough the center of of Energy:Ei=Efmgh=12mvcm2+12 Icm 2 (1)Since the disk rolls without slipping:8vcm=R =vcmR (2)(1): mgh=12mvcm2+1212mR2()vcmR 2=34mvcm2vcm=4gh3(2): =4gh3R2 The speed at the bottom does not depend onthe radius or the mass of the disk. The speed at the bottom is less than when thedisk slides down a frictionless ramp:v=2gh The Angular speed depends on the radius butnot the mass.

5 We can still apply conservation of energyeven though there is a friction force. Thefriction force cannot dissipate mechanicalenergy because it is a static friction. The point9in contact with the surface is at rest while incontact with the surface. A sphere or a hollow cylinder will reach thebottom with different speeds due to differentcenters of mass:Efsphere=12mvcm2+1225mR2()vcmR 2mgh=710mvcm2vcm=107ghEfhollowcyl=12mvcm 2+12mR2()vcmR 2mgh=mvcm2vcm=ghHThe sphere will reach the bottom first,followed by the disk and hollow result is independent of the mass andradius!Example:A sphere rolls down a ramp as shown. Theramp and the floor has friction to keep the ball10rolling without slipping until it reaches thesecond ramp which is frictionless. How high onthe second ramp does the sphere rise?hhfrictionless12 The sphere rolls down the 1st ramp andreaches some Angular speed at the bottom.

6 The sphere rolls across the floor with thesame Angular speed. The sphere slides up the ramp with the sameangular speed because there is no torque(friction) acting on the (1)Efloor=12mv2+I 2=12m 2R2+12 25mR2 2=710m 2R2 (2)Ef=mgh2+12I 2=mgh2+12 25mR2 2=mgh2+15m 2R2 (3)Apply conservation of energy to (1) and (2):11mgh1=710m 2R2 2R2=107gh1 (4)Apply conservation of energy to (1) and (3):mgh1=mgh2+15m 2R2gh1=gh2+15 2R2(4): gh1=gh2+27gh1h2=57h1 Angular Momentum : rotational analogue of linear Momentum must be defined with respect to some pointrrpp similar to torque, two calculation methods:L=rp =rpsin =rmvsin L=r p=(rsin )p=rmvsin 12 A vector quantity: r L =r r r p H r L is perpendicular to r r and r p Hdirection is defined by the right hand rule unit.

7 Kg m2/s A particle does not have to travel in a circle tohave Angular Momentum . A particletraveling in a straight line has angularmomentum relative to a particular point. If a point that the Angular Momentum isdefined relative to is along the momentumvector, then the Angular Momentum is zero(r =0). For a system of particles: r L =r L i =r L 1+r L 2+..+r L nNewton s Second Law:Translational: r F ext=dr p dt Rotational: r ext=dr L dt 13 The vector sum of all the torques acting on aparticle is equal to the rate of change of theangular Momentum . Both and L must be defined relative to thesame Momentum around a Fixed Axis:rrLv zAngular Momentum of a small element ofmass dm on the object: dLz=dLsin =r(dm)vsin =r (dm)v= r 2dmLz= r 2dm Lz=I 14 The component of the Angular momentumalong a fixed axis of rotation is just themoment of inertia times the Angular velocity .

8 We often just write:L=I Other components often cancel by symmetryConservation of Angular Momentum : If any component of the net external torqueon a system is zero, then the component ofthe Angular Momentum of the system alongthat axis is conserved. If a rotating object can some how changes itsmoment of inertia by internal forces, then theobject will spin faster or slower depending onwhether the moment of inertia decreases orincreases:Li=LfIi i=If f f=IiIf i15 Example:An ice skater would start spinning with theirarms extended away from the center of her body(the axis of rotation). As the ice skater pulls herarms tight to her body, the mass is now closer tothe axis of rotation, therefore the moment ofinertia has been reduced and the skater spinsfaster in order to conserve Angular :A uniform thin rod of length m and kg can rotate in a horizontal plane about avertical axis through its center.

9 The rod is at restwhen a g bullet traveling the horizontal planeis fired into one end of the rod. As viewed fromabove, the direction of the bullet s velocity makesan angle of 60o with the rod. If the bullet lodgesin the rod and the Angular velocity of the rod is10 rad/s immediately after the collision, what isthe magnitude of the bullet s velocity just beforethe impact?16axisL Conservation of Angular Momentum :Li=LfmbvbL2sin =(Ir+Ib) f=112mrL2+mbL2 2 f f=mbvbL2sin 112mrL2+mbL2 2=1290 m/sExample:Consider a person standing on a platformthat can rotate. The person is holding a wheelthat is spinning such that its Angular momentumis pointing upwards:Lw17Li=LwIf the person turns the wheel over so that theangular Momentum of the wheel is pointingdown, what happen to the motion of the man?LwThe person must start rotating because theangular Momentum of the person-wheel systemmust be conserved:Lf= Lw+LpLw= Lw+LpLp=2Lw