A uniform spherical shell of mass M = 3.0 kg and radius R = 13.0 cm rotates abou

Question

A uniform spherical shell of mass M = 3.0 kg and radius R = 13.0 cm rotates about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.96×10-3 kg m2 and radius r = 6.0 cm, and its attached to a small object of mass m = 2.0 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it has fallen a distance h = 0.7 m from rest: Use work - energy considerations.

Explanation / Answer

Here,

Is = (2/3)*3*0.13^2 = 0.0338 kgm²
Ip = 3.96*10^-3 kgm²

Energy:
½Is*ws² + ½Ip*wp² = 2*g*0.7 - 0.5*2*v^2

But ws= v/0.13 and wp = v/0.06, so

0.5*0.0338*(v/0.13)^2 + 0.5*3.96*10^-3*(v/0.06)^2 = 2*g*0.7 - 0.5*2*v^2

v = 2.32075 m/s

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