_lec32_ppt_ - Lecture 32 P112 Announcements Evaluation...

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Unformatted text preview: Lecture 32 Apr 11, 2007 P112 Announcements Evaluation Forms Agenda for today Work in Rotational Motion Gyroscopic motion Work in Rotational Motion Work done by a tangential force (constant torque) dW = Ftan Rd dW = z d 2 W = 1 z d 2 1 W = z ( )= z Work and Energy in Rotation Work equals the change in kinetic energy dW = W = z d 1 d I d =I d dt 2 1 W = I 2 2 2 1 I 2 2 1 = K Power in Rotation rate of doing work by torque equals power dW P= = dt P= z z d dt Example 1: Disk & String A string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). How fast is the disk spinning after the string has unwound? R F M Solution: Disk & String... The work done is W = The torque is = RF (since = 90o) The angular displacement is 2 rad/rev x 10 rev. So W = (.1 m)(10 N)(20 rad) = 62.8 J Disk & String... WNET = W = 62.8 J = K Recall that I for a disk about its central axis is given by: 1 I = MR 2 2 So K= 1 1 MR 2 2 2 2 R M =W = 4W = 2 MR 4 (62.8J ) = 792.5 rad/s 2 (.04kg)(.1) Example 2: Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the pull ? 1 2 (a) disk 1 (b) disk 2 (c) same F F Solution The work done on both disks is the same! W = Fd The change in kinetic energy of each will therefore also be the same since W = K. But we know So since I1 = I2 1 1 K= I 2 2 1 2 = 2 F d F Review... A freely moving particle has a definite angular momentum about any axis. If no torques are acting on the particle, its angular momentum will be conserved. In the example below, the direction of L is along the z axis, and its magnitude is given by LZ = pd = mvd. y x d m v Example: Bullet hitting stick 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 v1, and the final speed is v2. What is the angular speed (Ignore gravity) F of the stick after the collision? M m v1 D F D/4 v2 Li = mv1 D 4 = L f = mv 2 D +I 4 F A general expression for L of a system: For a system of particles we can write L= i ri pi = i miri v i Express position and velocity in terms of the center of mass: ri = Rcm + ri* vi = Vcm + v*i where ri* and v*i are the position and velocity measured in the CM frame. r* Rcm r A general expression for L of a system... So we can write: Expanding this: L= i L= i (R cm + r *i ) mi (Vcm + v *i ) r *i mi v *i i R cm miVcm + i R cm mi v *i + i r *i miVcm + Which becomes: L = R cm M tot Vcm + R cm i Lcm m i v *i + i m i r *i Vcm + i r * i m i v *i =MV*cm = 0 =MR*cm = 0 L* So finally we get the simple expression L = Lcm+ L* Where LCM = RCM PCMis the angular momentum of the CM and L* is the angular momentum about the CM. The total angular momentum of a system about a given axis is the sum of the angular momentum of the center of mass about this axis and the angular momentum about an axis through the center of mass. A general expression for L of a system... L = Lcm+ L* Picture it this way: y origin (axis) x m,I CM d v ^ L* = I k due to rotation about CM ^ LCM = mvd k due to movement of CM Turning the bike wheel. A student sits on the turntable holding a bicycle wheel that is spinning in the horizontal plane. He flips the rotation axis of the wheel 180o What will happen? Turning the bike wheel... Since there are no external torques acting on the student-turntable system, angular momentum is conserved. Initially: LINI = LW,I Finally: LFIN = LW,F + LS LW,I LS LW,I = LW,F + LS LW,F Gyroscopic Motion: Suppose you have a spinning gyroscope in the configuration shown below: If the left support is removed, what will happen?? support g pivot Gyroscopic Motion... It does not fall down, instead it precesses around its pivot axis ! This rather odd phenomenon can be easily understood using the simple relation between torque and angular momentum we derived earlier. pivot Gyroscopic Motion... The magnitude of the torque about the pivot is = mgd. The direction of this torque at the instant shown is out of the page (using the right hand rule). The change in angular momentum at the instant shown must also be out of the page! d = dL dt L mg pivot Gyroscopic Motion... Consider a view looking down on the gyroscope. The magnitude of the change in angular momentum in a time dt is dL = Ld . So where is the "precession frequency" L(t) dL d L(t+dt) top view ( pivot d dL =L dt dt L Gyroscopic Motion... So = dL =L dt = L In this example, = mgd and L = I : The direction of precession is given by applying the right hand rule to find the direction of and hence of dL/dt. d = mgd I L mg pivot Summary Angular momentum is non-intuitive ...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell.

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