DescentBB

Lou is trying to kill mice by swinging a clock of mass m attached to one end of a light (massless) rope of length L = 1.8 m hanging on a nail in the wall. The clock end of the rope is free to rotate around its other end in a vertical circle. Lou raises the clock until the rope is horizontal, and when mice peek their heads out from the hole to their den, he gives it an initial downward velocity v. The clock misses a mouse and continues on its circular path with just enough energy to complete the circle and bonk Lou on the back of his head, to the sound of cheering mice.



(a) What was the value of v?

(b) What was the clock's speed at the bottom of its swing?

I know it has something to do with energy conservation and centripital accel, ac = v^2 / r = 9.8 at 12 oclock, but i dont know how to find the velocities at 3 oclock and 6 oclock.

nope wrong answer

sorry

Well, it's a string, so at the top you have to have enough acc. too keep it taught. Ac = g From that, you get a V at the top. Then, find the change in energy in the three heights- E12 and E3/9, where E6 = 0. Then, find the deltaV's due to these E's, and add those V's individually to your first V and you have your two answers.

The energy at 6 oclock == 0???? If that is true, it would never get to 9 oclock, and therefore would never complete the loop

Will there be an airplane and a conveyor belt involved here at any time??

I think energy considerations alone are enough. If v0=0, since the potential energy is just MgL, the m will swing between the 3 and 9 position. At the bottom of the swing all of this potential is converted to kinetic energy = 1/2 mv^2. Set PE=KE and solve for v. Now if you want m to go all the way around, you're going to need at least double this v at the start to get it to go all the way from the 3 to the 9 position.

ccb056 wrote:The energy at 6 oclock == 0???? If that is true, it would never get to 9 oclock, and therefore would never complete the loop

I'm talking about potential energy- sorry I didn't express that well. Assume that PE, or your "ground" state is at 6 o'clock - that that point it would have all KE and no PE. Basically, the total energy in the system will never change. You can find a V at the top b/c you have to keep the string taught, and from that you can find a total energy at the top. (KE + PE) You can then take that energy and subtract the PE at each of the other staters, and convert the remaining KE into a speed.