A uniform spherical shell of mass M = 2.0 kg and radius R = 14.0 cm rotates abou

Question

A uniform spherical shell of mass M = 2.0 kg and radius R = 14.0 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 = 4.14×10-3 kg m2 and radius r = 6.0 cm, and its attached to a small object of mass m = 3.0 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 = 0.9 m from rest: Use work - 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)*2*.14² = 8.1 kgm²
Ip = 4.14 ×10-3 kgm²

Energy:
½Is*ws² + ½Ip*wp² = 3*g*.9 - ½*3*v²

But ws= v/.14 and wp = v/.06, so

½*8.1*(v/.14)² + ½*.210*(v/.06)² =3*g*.9 - ½*3*v²

solve for v

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