Ch 11 Rolling, Torque and Angular Momentum. Question 10 Problems: 3, 5, 9, 11, 17, 19, 23, 27, 31, 37, 41, 45, 53, 58, 69

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1 Ch 11 Rolling, Torque and Angular Momentum Question 10 Problems: 3, 5, 9, 11, 17, 19, 3, 7, 31, 37, 41, 45, 53, 58, 69

2 Rolling motion combination of translational motion and rotational motion. Chapter will cover objects that roll smoothly along a surface. The object does not slip or bounce along the sufrace. For rolling motion, the center of the object moves in a line parallel to the surface. Other points on the object do not follow straight paths. See figure 11- to show the different paths taken by the two different points on a rolling wheel.

3 Relation between arc length and angle of rotation. s = R When the object rotates through the angle, a point at a distance R from the rotation axis moves through a distance of s. v com v com s s

4 s= R The arc length s is the same as the distance that the wheel translates. The linear (translational) speed, v com, of the wheel is ds/. The angular speed of the wheel is So: ds v com d R R = d /

5 Rolling motion is the combination of purely rotational motion and purely translational motion. pure rotation + pure translation = rolling motion v com v com v= v com v com v com v com v com v= -v com +v com =0 The velocity of a point at the top of the rolling wheel is twice that of the center of the wheel. v top = (R) = ( R) = v com Fig 11-5 shows a bicycle wheel, the spokes are more blurred at the top. They are moving faster than the those at the bottom.

6 Kinetic Energy of Rolling As an object rolls, the point at the very bottom, the contact point with the surface, is instantaneously stationary. We will call this point P and we can treat the rolling as rotation about this point. K = ½ I P I P is the rotational inertia about the point P Parallel axis theorem says: I P = I com + MR

7 K K K 1 I 1 I 1 I P com com 1 MR 1 Mv com The kinetic energy of a rolling object comes from the rotational kinetic energy and translational kinetic energy.

8 Forces of rolling If a wheel rolls smoothly, there is no sliding at the contact point so there is no friction. However, if there is an external force that produces an acceleration, there will be an angular acceleration,. The acceleration will make the wheel want to slide at the contact point. Then a frictional force will act on the wheel to oppose the tendency to slide. If the wheel does not slide the force is static frictional force. If the wheel were to slide, the force would by kinetic frictional force. However, this would not be smooth rolling motion.

9 Direction of static frictional force. If the wheel, moving to the right, were to accelerate, the bottom of the wheel would want to move to the left compared to the surface. Thus the static friction force is to the right. If the same wheel were to slow down, the direction of the acceleration and angular acceleration would switch, and the static friction force will now be pointing to the left.

10 Rolling down a ramp The direction of the static friction force is the confusing part here. It points up along the ramp. If the wheel were to slide down the ramp, the friction opposing the sliding would be pointing up. N x Fg sin f s f s Mg sin = M a com Fg Fg cos

11 = I Only force on the wheel that produces torque is the friction. R f s = I com will need to make use of: a com = R (a is down the ramp, negative x-direction, but the wheel rolls counterclockwise so is positive) a com R So we can solve for f s : f s I com a R com a com g sin I 1 MR com

12 Yo-yo A Yo-yo is behaves similar to the wheel rolling down a ramp. 1) Instead of rolling down ramp of angle, the yo-yo follows an angle of 900with the horizontal. ) The yo-yo rolls down a string on an axle of radius R. R o R o T 3) Instead of friction, the tension slows the yo-yo. a com g I 1 MR0 com R F g

13 Angular Momentum linear momentum p = mv Rotational analogue is angular momentum. Occurs when momentum is offset from some reference point. r p m( r v) rmvsin where is the smallest angle between r and v. rp rmv r p r mv See fig O r p

14 Newton s nd Law in Angular Form Proof: For rotations d dp F net net net F net r d ma r a r m d v v a r m d v dr dv r m d v r m ) ( ) ( ) ( ) (

15 Angular Momentum of a System of Particles Just like we earlier added up the individual momentums of a system of particles, we can do the same for angular momentums. L i i dl dl i net d i The net external torque on a system of particles is equal to the time derivative of the total angular momentum of a system of particles.

16 For a rigid body, all the particles that make up the body have the same angular velocity. L=I To get the direction of L, use the right hand rule. L will be in the same direction as. Direction will be perpendicular to plane of rotation.

17 Conservation of Angular Momentum Just like linear momentum is always conserved, angular momentum is a conserved quantity. Can see from Newton s nd for rotations, if there is no net external torque, the angular momentum is constant. So for any isolated system: L i = L f I i i = I f f

18 Gyroscope A simple gyroscope is a rotating wheel that is on an axle where one end of the axle is on a support. When you let go of the axle, the gyroscope want to fall down because of gravity. The torque from the weight of the axle changes the angular momentum of the wheel. For a rapidly spinning gyroscope, the angular momentum is fixed by L=I. The torque will only change the direction of the angular momentum vector. Thus the gyroscope rotates in the horizontal plane. This rotation is called precession.

19 L=I dl dl = Mg r If the gyroscope precesses through an angle, d. L (t) L ( t ) d dl d d dl L ( Mgr) I Mgr I

20 Problems: 8, 14, 4, 6, 48, 61

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