A uniform spherical shell of mass M = 7.0 kg and radius R = 11.0 cm rotates abou

Question

A uniform spherical shell of mass M = 7.0 kg and radius R = 11.0 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 = 3.38×10-3kg m2 and radius r = 5.0 cm, and its attached to a small object of mass m = 5.0 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 = 1.5 m from rest: Use work - energy considerations.

Explanation / Answer

The moment of inertia of the spherical shell = 2/3(MR2)

So, the kinetic energy of the system when the block has descended by a height ‘h’ = ½(2/3(MR2))2 + ½(I2) + ½(mv2)

Now, we know that = v/r, substituting for in the above expression, we get

½(2/3(MR2))(v/R)2 + ½(I(v/r)2) + ½(mv2)

Since the system started from rest, this energy must be equal to the potential energy lost by the block = mgh = 5.0 * 9.8 * 1.5 = 73.5 J

Therefore, ½(2/3(MR2))(v/R)2 + ½(I(v/r)2) + ½(mv2) = 73.5

v2(7/3) + 0.68v2 + (5/2)v2 = 73.5

v2 = 73.5/5.5

which gives v = 3.66 m/s

So the speed of the object after it has fallen a distance h = 1.5 m from rest is 3.66 m/s

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