A uniform spherical shell of mass M = 5.00 kg and radius R = 0.620 m can rotate

Question

A uniform spherical shell of mass M = 5.00 kg and radius R = 0.620 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 = 0.0890 kg·m2 and radius r = 0.0770 m, and is attached to a small object of mass m = 4.50 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 0.885 m after being released from rest? Use energy considerations.

This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3times

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)*5*.620² = 1.2813 kgm²
Ip = .0.089 kgm²

Energy:
½Is*ws² + ½Ip*wp² = 4.5*g*0.885 - ½*4.5*v²

But ws= v/0.62 and wp = v/.077, so

½*1.2813*(v/0.62)² + ½*.089*(v/.077)² = 4.5*g*0.885 - ½*4.5*v²
Now solving for v

v=1.8484

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