A uniform spherical shell of mass M = 7.0 kg and radius R = 18.0 cm rotates abou

Question

A uniform spherical shell of mass M = 7.0 kg and radius R = 18.0 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 = 5.40×10-3 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 = 0.7 m from rest: Use work - energy considerations.

Explanation / Answer

Apply the Newton's second law, we get

mog - T1 = moa

so, T1 = mog - moa = (1)(9.8) - (1)a  = 9.8 - a

The net torque is zero when the system is in equilibrium.

T2(r) = I ( = a/r)

I for a spherical shell is 2/3MR2

T2r = (2/3)msr2a/r

Simplify

T2 = (2/3)msa = (2/3)(7)(a) = 4.7a

On the pulley, the differenxe in the tensions will cause a torque on it

(T1 - T2)r = Ia/r

(9.8 - a - 4.7a)(0.06) = (5.4 X 10-3)(a)/(0.06)

Solve for a, we get

0.588 - 0.342a = 0.09a

a = 1.36 m/s2

use the equation vf2 = vo2 + 2ad to solve for the velocity.

vf2 = (0) + (2)(1.36)(0.7) = 1.9m/s

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