Announcement 1. Dr. Baba Jain will be holding a help session on Friday, Oct. 10, at 3:30 pm in UPL 110. 2. Look up your mini scores on the CAPA system and compare with your records! 3. Dr. Frawley is responsible for mini 3. Results were not good (7.9 average). 4. Mini 4: Next Thursday. I will spend 15-20 minutes today on preparation. 1 Rotation (Chapter 9 of Tipler-Mosca) Figure 9-2 of Tipler-Mosca shows a disk, which rotates about a xed axis perpendicular to the disk and through its center. Rigid body: As the disk turns, the distance between any two particles that make up the disk remains xed. Consider a particle at distance r from the center and let be the angle measures counterclockwise from a xed reference line. When the disk rotates through and angular displacement d , measured in radians (rad), the particle moves through a circular arc of length ds = r |d | . 2 The time rate of change of the angle is the angular velocity d = . dt Question: A CD-ROM disk is rotating at 3000 revolutions per minute. What is the angular speed in radians per second? 1. 284 rad/s 2. 314 rad/s 3. 334 rad/s. Angular acceleration: d d2 = = 2 . dt dt Tangential velocity: d ds =r = r . vt = dt dt 3 Tangential acceleration: dvt d at = =r = r . dt dt Do not confuse this with the centripetal acceleration 2 vt (r )2 ac = = = r 2 . r r 4 Rotational Kinetic Energy 1 Krot = 2 2 mi vi i 1 = 2 i 1 mi (ri ) = 2 2 2 mi r i i 2 The quantity I= i 2 mi r i , where ri is the distance of particle i from the rotation axis, is called Moment of Inertia. The kinetic energy becomes 1 Krot = I 2 . 2 For a continuous object: I= r2 dm . 5 Calculating the Moment of Inertia Discrete Systems of Particles: Simply apply the de nition I= i 2 mi r i . Examples (PRS): For gure 9-3 of Tipler-Mosca the result is I = n m a2 where n is an integer. Push this number. For gure 9-4 of Tipler-Mosca n is another (or the same?) integer. Push this number. 6 Continuous Objects: The moment of inertia is calculated by integration. Examples folow. 1. Uniform stick about an axis perpendicular to the stick and through one end ( gure 9-5 of Tipler-Mosca). L L I= 0 x dm = 0 2 M x dx = L 2 pick one! 1 1. M L3 3 1 2. M L2 3 1 3. M L2 . 2 7 2. Hoop of radius R about a perpendicular axis through its center ( gure 9-6 of Tipler-Mosca). I= r2 dm = R2 dm = M R2 . 3. Uniform disk of radius R about a perpendicular axis through its center ( gure 9-7 Tipler-Mosca). M dAr dm = A so that R R where A = R2 and Ar = r2 dAr = 0 0 2 r dr = 2 R r 0 = R2 = A . 8 Therefore, and M 2M dm = 2 r dr = 2 r dr 2 R R I= 2M 2 r dm = 2 R R r3 dr = 0 pick one! 1 1. M R4 4 1 2. M R4 2 1 3. M R2 2 2 4. M R2 . 3 9 3. Uniform cylinder of radius R about its axis ( gure 9-8 of TiplerMosca). Using the result for the uniform disk, we have 1 2 I= R 2 1 dmdisk = M R2 . 2 Results for Uniform Bodies of Various Shapes are collected in Table 9-1 of Tipler-Mosca! Examples we had: hoop I = M R , 2 1 full disk I = M R2 . 2 10 Rolling Objects without Slipping The point of contact moves a distance s = R . The velocity and acceleration of the CM are vcm = R and acm = R . The kinetic energy of a rolling object is 1 1 2 K = M vc...