Physics 1501: Lecture 22, Pg 1 Physics 1501: Lecture 22 Today’s Agenda l Announcements çHW#8: due Oct. 28 l Honors’ students çsee me Wednesday at 2:30 in P-114 l Topics çRolling motion çAngular Momentum çFigure Skating Day !!! + other fun demos …

Physics 1501: Lecture 22, Pg 2 Torque, Work, Kinetic Energy  r F We can define torque as:  = r x F = r F sin    X = y F Z - z F Y  Y = z F X - x F Z  Z = x F Y - y F X We find the work : W =  Kinetic Energy of rotation: K = ½ I   Recall the Work Kinetic-Energy Theorem:  K = W NET l So for an object that rotates about a fixed axis: rF x y z  

Physics 1501: Lecture 22, Pg 3 Connection with CM motion... l So for a solid object which rotates about its center of mass and whose CM is moving:  V CM

Physics 1501: Lecture 22, Pg 4 Rolling Motion l Cylinders of different I rolling down an inclined plane: h v = 0  = 0 K = 0 R  K = -  U = Mgh v =  R M

Physics 1501: Lecture 22, Pg 5 Rolling... Use v=  R and I = cMR 2. So: The rolling speed is always lower than in the case of simple sliding since the kinetic energy is shared between CM motion and rotation. hoop: c=1 disk: c=1/2 sphere: c=2/5 etc... c c c c

Physics 1501: Lecture 22, Pg 6 Lecture 22, ACT 1 Rolling Motion l A race !! Two cylinders are rolled down a ramp. They have the same radius but different masses, M 1 > M 2. Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie M1  h M? M2 Active Figure

Physics 1501: Lecture 22, Pg 7 Example : Rolling Motion l A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M  h M v ? Ball has radius R M M M M M

Physics 1501: Lecture 22, Pg 8 Example : Rolling Motion l Use conservation of energy. E i = U i + 0 = Mgh E f = 0 + K f = 1/2 Mv 2 + 1/2 I  2 = 1/2 Mv 2 + 1/2 (1/2MR 2 )(v/R) 2 Mgh = 1/2 Mv 2 + 1/4 Mv 2 v 2 = 4/3 g h v = ( 4/3 g h ) 1/2

Physics 1501: Lecture 22, Pg 9 Consider a roller coaster. We can get the ball to go around the circle without leaving the loop. Note: Radius of loop = R Radius of ball = r Example : Roller Coaster

Physics 1501: Lecture 22, Pg 10 How high do we have to start the ball ? Use conservation of energy. Also, we must remember that the minimum speed at the top is v top = (gR) 1/2 E 1 = mgh E 2 = mg2R + 1/2 mv 2 + 1/2 I  2 = 2mgR + 1/2 m(gR) + 1/2 (2/5 mr 2 )(v/r) 2 = 2mgR + 1/2 mgR + (2/10)m (gR) = 2.7 mgR 1 2

Physics 1501: Lecture 22, Pg 11 How high do we have to start the ball ? E 1 = mgh E 2 = 2.7 mgR mgh = 2.7 mgR h = 2.7 R h = 1.35 D (The rolling motion added an extra 2/10 R to the height: without it, h = 2.5 R) 1 2

Physics 1501: Lecture 22, Pg 12 Chap. 11: Angular Momentum When we write  = I  we are really talking about the z component of a more general vector equation. (we normally choose the z-axis to be the the rotation axis.)  z = I z  z l We usually omit the z subscript for simplicity. z zz zz IzIz

Physics 1501: Lecture 22, Pg 13 Angular Momentum : angular momentum r m F l An object of mass m is rotating in a circular path under the action of a constant torque:

Physics 1501: Lecture 22, Pg 14 Angular Momentum l The torque acting on an object is equal to the time rate of change of the object’s angular momentum l The angular momentum of an system is conserved when the net external torque acting on the system is zero. That is, when  = 0, the initial angular momentum equals the final angular momentum. L f = L i

Physics 1501: Lecture 22, Pg 15 Angular Momentum: Definitions & Derivations l We have shown that for a system of particles Momentum is conserved if l What is the rotational version of this ?? F  The rotational analogue of force F is torque  p l Define the rotational analogue of momentum p to be angular momentum p=mv

Physics 1501: Lecture 22, Pg 16 Definitions & Derivations... L l First consider the rate of change of L: So(so what...?)

Physics 1501: Lecture 22, Pg 17 Definitions & Derivations... l Recall that l Which finally gives us: l Analogue of !!   EXT

Physics 1501: Lecture 22, Pg 18 What does it mean? l where and l In the absence of external torques Total angular momentum is conserved Active torqueActive angular momentum

Physics 1501: Lecture 22, Pg 19 i j Angular momentum of a rigid body about a fixed axis: l Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle: rr1rr1 rr3rr3 rr2rr2 m2m2 m1m1 m3m3  vv2vv2 vv1vv1 vv3vv3 L We see that L is in the z direction. Using v i =  r i, we get L =I (since r i, v i, are perpendicular) Analogue of p = mv !! 

Physics 1501: Lecture 22, Pg 20 Angular momentum of a rigid body about a fixed axis: In general, for an object rotating about a fixed (z) axis we can write L Z = I  The direction of L Z is given by the right hand rule (same as  ). We will omit the ” Z ” subscript for simplicity, and write L = I   z L Z = I 

Physics 1501: Lecture 22, Pg 21 Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity  0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity  F. 00 z FF z

Physics 1501: Lecture 22, Pg 22 Example: Two Disks l First realize that there are no external torques acting on the two-disk system. çAngular momentum will be conserved ! l Initially, the total angular momentum is due only to the disk on the bottom: 00 z 2 1

Physics 1501: Lecture 22, Pg 23 Example: Two Disks l First realize that there are no external torques acting on the two-disk system. çAngular momentum will be conserved ! l Finally, the total angular momentum is due to both disks spinning: FF z 2 1

Physics 1501: Lecture 22, Pg 24 Example: Two Disks l Since L INI = L FIN 00 z FF z L INI L FIN An inelastic collision, since E is not conserved (friction) ! 1 / 2 MR 2    MR 2  F  F = 1 / 2   

Physics 1501: Lecture 22, Pg 25 Demonstration of Conservation of Angular Momentum l Figure Skating : AA z BB z Arm I A < I B  A >  B L A = L B

Physics 1501: Lecture 22, Pg 26 Angular Momentum Conservation l A freely moving particle has a definite angular momentum about any given axis. l If no torques are acting on the particle, its angular momentum will be conserved. L l In the example below, the direction of L is along the z axis, and its magnitude is given by L Z = pd = mvd. y x v d m

Physics 1501: Lecture 22, Pg 27 Example: Bullet hitting stick l A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v 1, and the final speed is v 2.  What is the angular speed  F of the stick after the collision? (Ignore gravity) v1v1 v2v2 M FF beforeafter m D D/4

Physics 1501: Lecture 22, Pg 28 Example: Bullet hitting stick... l Conserve angular momentum around the pivot (z) axis! l The total angular momentum before the collision is due only to the bullet (since the stick is not rotating yet). v1v1 D M before D/4m

Physics 1501: Lecture 22, Pg 29 Example: Bullet hitting stick... l Conserve angular momentum around the pivot (z) axis! l The total angular momentum after the collision has contributions from both the bullet and the stick.  where I is the moment of inertia of the stick about the pivot. v2v2 FF after D/4

Physics 1501: Lecture 22, Pg 30 Example: Bullet hitting stick... l Set L BEFORE = L AFTER using v1v1 v2v2 M FF beforeafter m D D/4

Physics 1501: Lecture 22, Pg 31 Example: Throwing ball from stool A student sits on a stool which is free to rotate. The moment of inertia of the student plus the stool is I. She throws a heavy ball of mass M with speed v such that its velocity vector passes a distance d from the axis of rotation.  What is the angular speed  F of the student-stool system after she throws the ball ? top view: before after d v M I I FF

Physics 1501: Lecture 22, Pg 32 Example: Throwing ball from stool... l Conserve angular momentum (since there are no external torques acting on the student-stool system):  L BEFORE = 0, L stool = I  F  L AFTER = 0 = L stool - L ball, L ball = I ball  ball = Md 2 (v/d) = M d v  0 = I  F - M d v top view: before after d v M I I FF  F = M v d / I 