105Sol3 - Physics 105 Solutions to Problem Set 3 By Stephen...

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Physics 105 Solutions to Problem Set 3 By Stephen Glassman, Fall 2007 4.2 (a) We evaluate the integral by breaking up the path at Q : W = P O F · dr = Q O F · dr + P Q F · dr = Q O F x dx + Q O F y dy + P Q F x dx + P Q F y dy = 1 0 x 2 dx + 0 0 0 dy + 1 1 dx + 1 0 2 ydy = 1 3 + 0 + 0 + 1 = 4 3 (b) We follow the suggestion in the text: W = P O F · dr = P O F x dx + P O F y dy = P O x 2 dx + P O 2 xydy = P O x 2 dx + P O 4 x 4 dx = 1 0 ( x 2 + 4 x 4 ) dx = 1 3 + 4 5 = 17 15 (c) As t ranges from 0 to 1, our path goes from O to P . This time x = t 3 , dx = 3 t 2 dt , y = t 2 , and dy = 2 tdt . 1
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W = P O F · dr = P O F x dx + P O F y dy = P O x 2 dx + P O 2 xydy = P O 3 t 8 dt + P O 4 t 6 dt = 1 0 (3 t 8 + 4 t 6 ) dt = 3 9 + 4 7 = 57 63 Note that our work is path-dependent and therefore our force is not conservative. 4.4 (a) Because the string exerts no torque, angular momentum is conserved. We can calculate the angular momentum using L = r × p . For circular motion, we recall r p and v = ωr . Therefore, mr 2 ω = L = mr 2 0 ω 0 and so ω = ω 0 ( r 0 r ) 2 . (b) We can draw the string in so long as we pull with a force infinitesimally greater than the centripetal force. So in the limit that we pull slowly, the tension remains approximately the centripetal force. However, this tension now does work since the particle moves radially. We calculate this work: W = P O F · dr = r r 0 - m ( ω ( r )) 2 r 2 dr = r r 0 - m ω 0 r 2 0 r 2 2 r dr = - 2 0 r 4 0 r r 0 ( r - 3 ) dr = - 2 0 r 4 0 - 1 2 r - 2 r r 0 2
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= 1 2 2 0 r 4 0 1 r 2 - 1 r 2 0 (c) Recall from the first problem set that v = ωr for uniform circular motion. Then: Δ KE = 1 2 m Δ( v 2 ) = 1 2 m Δ( ω 2 r 2 ) = 1 2 m ( ω 2 r 2 - ω 2 0 r 2 0 ) = 1 2 m ω 0 r 2 0 r 2 2 r 2 - ω 2 0 r 2 0 = 1 2 2 0 r 4 0 1 r 2 - 1 r 2 0 Not surprisingly, W = Δ KE . Suppose instead that we pull with a constant extra of tension. This results in an extra contribution of Δ r in our work calculation. We might ask: where does this work go? Well, if we don’t draw the string in slowly, then the mass arrives at the inner circle with non-zero radial velocity. This extra work is precisely that radial contribution to the final kinetic energy. Now when we stop pulling on the string, the mass continues on its inward trajectory. The string goes slack for a moment as the mass cuts across a chord of the inner circle. What precisely happens when the mass reaches the end of the chord will depend sensitively on the precise construction of the system. A likely possibility is that the abrupt stretching of the string will convert the radial contribution to kinetic energy into heat in the manner of inelastic collisions. By the way, we could also analyze this system in the (instantaneous) co-rotating frame. What force is responsible for speeding up the mass’ rotation as we draw in the string in this frame? It’s the coriolis force F cor = 2 m ˙ r × ω ! Let’s solve for ω ( r ): = τ = mr 2 ˙ ω = - 2 mr ˙ = ˙ ω ω = - 2 ˙ r r = ω ω 0 ˙ ω ω dt = - 2 r r 0 ˙ r r dt 3
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= ω ω 0 ω = - 2 r r 0 dr r = ln ω ω 0 = - 2 ln r r 0 = ω = ω 0 r 0 r 2 which is precisely our result from part a !
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