A uniform spherical shell of mass M = 5.5 kg and radius R = 14.0 cm rotates abou

Question

A uniform spherical shell of mass M = 5.5 kg and radius R = 14.0 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 = 4.86×10-3 kg m2 and radius r = 6.0 cm, and its attached to a small object of mass m = 3.0 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

Using work energy theorem

Work done by net force = change in kinetic energy

Wnet = K2-K1

Wnet = Work done by gravitational force = m*g*h

K2 = 0.5*I1*w^2 + 0.5*I2*w^2+ (0.5*m*v^2)

I1 = (2/5)*M*R^2 = (2/5)*5.5*0.14^2 = 0.04312 kg-m^2

w = V/R

I2 = 4.86*10^-3 kg-m^2

m = 3 kg

v = ?

then

m*g*h = (0.5*I1*w^2)+(0.5*I2*w^2)+(0.5*m*v^2)

(3*9.8*1.5) = (0.5*0.04312*(V/0.14^2)) + (0.5*4.86*10^-3*(V/0.06^2))+(0.5*3*v^2)

v = 3.67 m/s

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