A tetherball of mass m is rotating around a pole so that the rope attaching it i

Question

A tetherball of mass m is rotating around a pole so that the rope attaching it is getting shorter. The radius of rotation changes at rate of 0.1m/s. If the rope is initially 2m long, how long (how much time) will it take for the rate of rotation to double?

Explanation / Answer

I think 10 seconds. r = (2 - 0.1t)e^iwt dr/dt = v = 2iwe^iwt - 0.1iwte^iwt - 0.1e^iwt w = v/r = (2iw - 0.1iwt - 0.1)/ (2 - 0.1t) take the real part and w = -0.1/(2-0.1t) --> 2w - 0.1wt = -0.1 --> 1/w = t - 20 so when t=0 --> 1/w = -20 --> when t = 10 1/w = -10. This also makes sense because when t = 20, w = 0 as it should when the radius is 0. Also as t becomes larger w must clearly also become larger so this equation makes qualitative sense, I'm pretty sure it's correct.

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