_lec32_ppt_ - Lecture 32 P112 Announcements Evaluation...

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Lecture 32 Apr 11, 2007 P112
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Announcements ± Evaluation Forms
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Agenda for today ± Work in Rotational Motion ± Gyroscopic motion
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Work in Rotational Motion ± Work done by a tangential force (constant torque) dW = F tan Rd ± dW = ² z d ± W = ² z d ± ± 1 ± 2 ³ W = ² z ( ± 2 ´ ± 1 ) = ² z µ ±
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Work and Energy in Rotation ± Work equals the change in kinetic energy dW = ± z d ² W = I d ³ dt d ² ´ = I d ³ ³ 2 ³ 1 ´ µ ³ W = 1 2 I ³ 2 2 1 2 I ³ 1 2 = · K
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Power in Rotation ± rate of doing work by torque equals power P = dW dt = ± z d ² dt P = ± z ³
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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? F R M
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Solution: Disk & String... ± The work done is W = ± ² ± The torque is ± = RF (since ³ = 90 o ) ± The angular displacement ² is 2 ´ rad/rev x 10 rev. ± So W = (.1 m)(10 N)(20 ´ rad) = 62.8 J
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Disk & String... W NET = W = 62.8 J = ± K Recall that I for a disk about its central axis is given by: So ² = 792.5 rad/s ² R M ± = 4 W MR 2 = 4 62.8 J ( ) .04 kg ( ) .1 ( ) 2 ± K = 1 2 1 2 MR 2 ² ³ ´ µ · ¸ 2 = W I = 1 2 MR 2
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Example 2: Work & Energy ±
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