A uniform spherical shell of mass M = 15.0 kg and radius R = 0.900 m can rotate about a vertical axis on frictionless bearings (Fig. 10-44). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 0.210 kg·m2 and radius r = 0.130 m, and is attached to a small object of mass m = 2.40 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 1.47 m after being released from rest? Use energy considerations.
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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)*15*.900² = 8.1 kg∙m²
Ip = .210 kg∙m²
Energy:
½Is*ws² + ½Ip*wp² = 2.40*g*1.47 - ½*2.40*v²
But ws= v/.900 and wp = v/.130, so
½*8.1*(v/.9)² + ½*.210*(v/.130)² = 2.40*g*1.47 - ½*2.4*v²
I'm sure you can solve this for v.......
I got 1.669 m/s
Is = (2/3)*15*.900² = 8.1 kg∙m²
Ip = .210 kg∙m²
Energy:
½Is*ws² + ½Ip*wp² = 2.40*g*1.47 - ½*2.40*v²
But ws= v/.900 and wp = v/.130, so
½*8.1*(v/.9)² + ½*.210*(v/.130)² = 2.40*g*1.47 - ½*2.4*v²
I'm sure you can solve this for v.......
I got 1.669 m/s