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Welcome back to Physics 211

Welcome back to Physics 211. Today’s agenda (Tue Dec 7 th ): Conservation of Angular Momentum. An Equivalent definition of Angular Momentum. Consider a particle of mass m i at a distance r i from the center of rotation O

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Welcome back to Physics 211

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  1. Welcome back to Physics 211 Today’s agenda (Tue Dec 7th): Conservation of Angular Momentum

  2. An Equivalent definition of Angular Momentum • Consider a particle of mass mi at a distance ri from the center of rotation O • It is performing circular motion about O so that its speed vi is equal of riw • Hence its angular momentum Li is Li = mi vi ri = miw ri2

  3. Angular Momentum in terms of Moment of Inertia • The sum total angular momentum L is ∑ Li = ∑ miri2w = I w Here I is the moment of inertia about O • We thus arrive at the useful relation L = I w

  4. Newton’s Second Law for Rotation in terms of Angular Momentum Back to slide on rotational dynamics: miri2Dw/Dt = ti Rewrite: using Li=miri2w DLi/ Dt= ti Summing over all particles in body DL/ Dt=text L is the angular momentum=Iw

  5. Demo • Rotating stand plus dumbbells – spin faster when arms drawn in • Explanation – Refer to textbook page number 384

  6. QUESTIONAn ice skater spins about a vertical axis through her body with her arms held out. As she drawsher arms in, her angular velocity • 1. increases • 2. decreases • 3. remains the same • 4 need more information

  7. ANSWER • L = I w is conserved • As she draws her arms in, the radius r decreases, which leads to a decrease in I. • Since L has to be conserved, w has to increase

  8. Bicycle wheel demo • Spin wheel – generates L along axis of rotation. • If hold by axle – no torque applied. • Then L must remain fixed in magnitude (constant angular velocity) and direction • Hard to point in other directions (basis of gyroscope for navigation)

  9. Bicycle Wheel Demo Torque Applied by Hand • You stand on a turntable. • Plane of wheel is horizontal • Rotation is counterclockwise, as seen from above. • Then, you turn the wheel upside down, so it is rotating clockwise

  10. Question • Will the table rotate? • If so, in what direction? • Use conservation of angular momentum Ltotal. • Choose the system to be the wheel plus person plus turntable.

  11. Explanation • This system can be regarded as isolated. • The external torques on it are very small. • Direction of Lwheel is upward, initially. • Initially, the table has no rotation. • Then, Ltotal = Lwheel and is upward.

  12. Application of Conservation Law • Since the total angular momentum is constant, the final angular momentum of the system must remain in the upward direction. • But the final angular momentum of the wheel is downward. • Hence the turn table must have its angular momentum upward, to keep ∆Ltotal = 0.

  13. Conclusion & Observation • The turn table must turn counter-clockwise in order for its angular momentum to be upward. • This is what we observe

  14. Combining linear and rotational motion • Sometimes it is easy to interrelate the linear and rotational motion • rolling without slipping

  15. Example: Object rolling down incline • A disk of mass m rolls down an incline. • We desire the translational acceleration aCM of the disk. • We are given the inclination angle theta • The moment of inertia I of the disk about its center and the radius r of the disk.

  16. Free body diagram • There are three forces on the disk. • The weight force W acting downward, the normal force N perpendicular to the incline and the friction force F directed up the incline. • We apply the law for the translational acceleration : m aCM = FNET • We also apply the rotational law τ= Ia about the center of the disk.

  17. Rolling without slipping N F Wsin(q)-F=maCM FR=Ia Now aCM=Ra if no slipping  Wsin(q)-F=m aCM and F= I aCM/R2 W q (m+I/R2) aCM= Wsin(q)

  18. Summary • The rolling motion obeys the usual form of Newton’s second law if we replace the true mass m with m+I/R2 on the left hand side of our major result which is • (m+I/R2) aCM= W sin(θ) • Using the relation W = mg and solving for aCM we obtain aCM = mg sin(θ) /(m+I/R2). • Suppose the mass is in the form of a hoop of radius R. • Then all the mass is concentrated on the rim. It follows that the moment of inertia is simply I = mR2 • The result for the center of mass acceleration becomes simply aCM = (½)g sin(θ)

  19. Continued Summary • Note that for sliding motion, (no rotation) the result for the acceleration is aCM = g sin(θ). • Thus for a hoop, the acceleration for rolling motion is one-half that of sliding motion. • Important note: We used the relationship aCM=Ra This applies to any motion of rolling without slipping. Refer to the textbook page number 375 for justification.

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