A uniform spherical shell of mass M = 7.30 kg and radius R = 0.640 m can rotate

Question

A uniform spherical shell of mass M = 7.30 kg and radius R = 0.640 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 = 0.0980 kg·m2 and radius r = 0.130 m, and is attached to a small object of mass m = 3.10 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 1.19 m after being released from rest? Use energy considerations.

Explanation / Answer

let

v is the speed of hanging object when it falls 1.19 m.

Apply

gain in kinetic energy = loss of potentail energy

0.5*I_sphere*w_sphere^2 + 0.5*I_pulley*w_pulley^2 = m*g*h

0.5*(2/3)*M*R^2*w_sphere^2 + 0.5*I_pulley*(v/r)^2 = m*g*h

(1/3)*M*v^2 + 0.5*I_pulley*v^2/r^2 = m*g*h

v^2*( M/3 + 0.5*I_pulley/r^2) = m*g*h

v^2*(7.3/3 + 0.5*0.098/0.13^2) = 3.1*9.8*1.19

v^2*5.33 = 36.15

v = sqrt(36.15/5.33)

= 2.6 m/s <<<<<<<<<<<<<<<<<<<----------------Answer

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