A uniform spherical shell of mass M = 11.0 kg and radius R = 0.840 m can rotate

Question

A uniform spherical shell of mass M = 11.0 kg and radius R = 0.840 m can rotate 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 = 0.0870 kg·m2 and radius r = 0.0770 m, and is attached to a small object of mass m = 3.30 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 when it has fallen a distance 1.15 m after being released from rest? Use energy considerations.

Explanation / Answer

The sum of the kinetic energies in the 2 rotating items will equal the loss in PE of the mass - the KE of the mass:

Is = (2/3)*11*0.840² = 5.174 kgm²
Ip = 0.087 kgm²

Energy:
½Is*ws² + ½Ip*wp² = 3.30*g*1.15 - ½*3.3*v²

But ws= v/0.840 and wp = v/0.0770, so

½*5.174*(v/0.84)² + ½*0.087*(v/0.077)² = 3.30*g*1.15 - ½*3.3*v²
3.67 v² + 7.34 v² = 37.2 – 1.65 v²

Solving, v = 1.714 m/s

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