A uniform spherical shell of mass M = 1.0 kg and radius R = 10.0 cm rotates abou

Question

A uniform spherical shell of mass M = 1.0 kg and radius R = 10.0 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 = 4.68×10-3 kg m2 and radius r = 6.0 cm, and its attached to a small object of mass m = 1.0 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 = 1.2 m from rest: Use work - energy considerations.

Explanation / Answer

Here, I_sphere = (2/5) m (r_sphere)² = (2/5) 1 (0.10)²

I_sphere = 0.004 kg m²

conserve of energy

total initial energy = total final energy

potential energy = rotational kinetic energy of sphere + rotational kinetic energy of pulley + translation kinetic energy of object

m g h = ½ (I_sphere) (_sphere)² + ½ (I_pulley) (_pulley)² + ½ m v²

m g h = ½ (I_sphere) (v/r_sphere)² + ½ (I_pulley) (v/r_pulley)² + ½ m v²

2(9.8)(1.2) = ½ (0.004)(v/0.10)² + ½ (4.68 E-3) (v/0.06)² + ½ (1) v²

=> 23.52 = v²[ 0.20 + 0.65 + 0.5] =  v²[1.35]

=> v = ± 4.17 m/s

choose positive value, so speed of the object is 4.17 m/s downward

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