A massless string is wrapped around the equator of a solid sphere (mass M = 53.5

Question

A massless string is wrapped around the equator of a solid sphere (mass M = 53.5 kg, radius R = 0.263 m). Mary holds the free end of the string, and the sphere is released from rest, Assume:
- the sphere is always parallel to the floor
- the string is always perpendicular to the radius of the sphere
- the string does not slip over the sphere

Suppose now that Mary holds the end of the string stationary, and the string unwinds as the sphere falls. Find:
- T, the magnitude of the tension in the string as the sphere turns: 4 N
- t, the time it takes for the falling sphere to reach an angular speed ? = 27.2 rad/s: 5 s

Explanation / Answer

mg- T = ma
Tr= I = I * a/r

I =2/5 mr2

=> T = 2/5ma

=> a = 5/7 g

=> T = 0.4* 53.5 * 5/7* 9.81= 149.95 N

= a/r = 5g/7r

t = / = 27.2* 7*0.263/(5*9.81) = 1.02 seconds

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