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

• Notes
• 13

This preview shows pages 1–5. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

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
= 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

This preview has intentionally blurred sections. Sign up to view the full version.

= 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 !
This is the end of the preview. Sign up to access the rest of the document.
• Fall '07
• EdgarKnobloch

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern