A uniform solid ball has a few turns of light string wound aroundit. If the end

Question

A uniform solid ball has a few turns of light string wound aroundit. If the end of the string is held steady and the ball is allowedto fall under gravity, what is the acceleration of the center ofthe ball? (Assume the string remains vertical)

I worked it out and found the answer to be 5/7g but I'm not sure ifits right. I got this by assuming the ball rotated around itscenter of mass so I used 2/5mr2 as the moment of inertiabut I wonder if I should have used the parallel axis theorem(I =Ml2 + Icm) for the moment of inertia instead,taking "l" as the distance between the point where the string isattached and the center of mass of the ball.

Explanation / Answer

You are right, or do like this: mg - T = ma T*r = (2mr2/5)*(a/r), so T = 2ma/5 mg - 2ma/5 = ma mg = 7ma/5 a = 5g/7

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