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Welcome back to Physics 215

Welcome back to Physics 215. Today ’ s agenda: More on angular momentum Rolling without slipping Review. Current homework assignment. HW10: Knight Textbook Ch.12: 70, 76, 88 2 exam-style problems from course website due Wednesday, Nov. 16 th in recitation.

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Welcome back to Physics 215

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  1. Welcome back to Physics 215 Today’s agenda: More on angular momentum Rolling without slipping Review

  2. Current homework assignment • HW10: • Knight Textbook Ch.12: 70, 76, 88 • 2 exam-style problems from course website • due Wednesday, Nov. 16th in recitation

  3. Exam 3: Thursday (11/17/11) • Material covered: • Textbook chapters 10-12 • Lectures up to and including 11/15 (slides online) • Workshop Activities • Homework assignments • As with Exams 1 & 2, Exam 3 is closed book, but you may bring calculator and onehandwritten 8.5” x 11” sheet of notes • Practice versions of Exam 3 posted online

  4. Angular Momentum 1. q r p O Point particle: |L| = |r||p|sin(q) = m|r||v| sin(q) vector form  L = r x p – direction of L given by right hand rule (into paper here) L = mvr if v is at 900 to r for single particle

  5. Angular Momentum 2. w o rigid body: * |L| = Iw(fixed axis of rotation) * direction – along axis – into paper here

  6. Rotational Dynamics • t=Ia DL/Dt = t • These are equivalent statements • If no net external torque: t=0  • * L is constant in time • * Conservation of Angular Momentum • * Internal forces/torques do not contribute • to external torque.

  7. Force Acceleration Momentum Kinetic energy Torque Angular acceleration Angular momentum** Kinetic energy Linear and rotational motion

  8. A hammer is held horizontally and then released. Which way will it fall?

  9. General motion of extended objects • Net force  acceleration of CM • Net torque about CM  angular acceleration (rotation) about CM • Resultant motion is superposition of these two motions • Total kinetic energy K = KCM + Krot

  10. Three identical rectangular blocks are at rest on a flat, frictionless table. The same force is exerted on each of the three blocks for a very short time interval. The force is exerted at a different point on each block, as shown. After the force has stopped acting on each block, which block will spin the fastest? 1. A. 2. B. 3. C. 4. A and C. Top-view diagram

  11. Three identical rectangular blocks are at rest on a flat, frictionless table. The same force is exerted on each of the three blocks for a very short time interval. The force is exerted at a different point on each block, as shown. After each force has stopped acting, which block’s center of mass will have the greatest speed? 1. A. 2. B. 3. C. 4. A, B, and C have the same C.O.M. speed. Top-view diagram

  12. Rolling without slipping translation rotation vcm = acm =

  13. Rolling without slipping N F = maCM  = Ia Now aCM = Ra if no slipping So, maCM and F = F W q

  14. A ribbon is wound up on a spool. A person pulls the ribbon as shown. Will the spool move to the left, to the right, or will it not move at all? 1. The spool will move to the left. 2. The spool will move to the right. 3. The spool will not move at all.

  15. A ribbon is wound up on a spool. A person pulls the ribbon as shown. Will the spool move to the left, to the right, or will it not move at all? 1. The spool will move to the left. 2. The spool will move to the right. 3. The spool will not move at all.

  16. A ribbon is wound up on a spool. A person pulls the ribbon as shown. Will the spool move to the left, to the right, or will it not move at all? 1. The spool will move to the left. 2. The spool will move to the right. 3. The spool will not move at all.

  17. Spinning a cylinder Cable wrapped around cylinder. Pull off with constant force F. Suppose unwind a distance d of cable 2R F • What is final angular speed of cylinder? • Use work-KE theorem • W = Fd = Kf = (1/2)I2 • Mom. of inertia of cyl.? -- from table: (1/2)mR2 • from table: (1/2)mR2 •  = [2Fd/(mR2/2)]1/2 = [4Fd/(mR2)]1/2

  18. cylinder+cable problem -- constant acceleration method extended free body diagram N F * no torque due to N or FW * why direction of N ? * torque due to t= FR * hence a= FR/[(1/2)MR2] = 2F/(MR)  wt = [4Fd/(MR2)] 1/2 FW radius R  = (1/2)t2 = d/R; t = [(MR/F)(d/R)]1/2

  19. Tabulated Results for Moments of Inertia of some rigid, uniform objects (from p.299 of University Physics, Young & Freedman)

  20. Reading assignment • Prepare for Exam 3 ! • Ch. 13 -- Gravitation (after Thanksgiving)

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