Question

A uniform spherical shell of mass M = 3.2 kg and radius R = 7.8 cm can rotate about a vertical axis on frictionless bearings (see figure below). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 ? 10^?3 kg · m² and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 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 90 cm after being released from rest? Use energy considerations.

--m/s

Answer 1

given
M = 3.2 kg
R = 7.8 cm = 0.078 m
Moment of inertia of shell, I_shell = (2/3)*M*R^2

= (2/3)*3.2*0.078^2

= 0.01297 kg.m^2

I_pulley = 3*10^-3 kg.m^2

r = 5 cm = 0.05 m

m = 0.6 kg

h = 90 cm 0.9 m

let v is the speed of falling mass, w1 is the angular speed of shell and w2 is the angular speed of the pulley.

v = R*w1 ===> w1 = v/R1 = v/0.078
v = r*w2 ===> w2 = v/r = v/0.05

Apply conservation of energy

gain in kinetic energy = loss of potential energy

(1/2)*m*v^2 + (1/2)*I_shell*w1^2 + (1/2)*I_pulley*w2^2 = m*g*h

(1/2)*0.6*v^2 + (1/2)*0.01297*(v/0.078)^2 + (1/2)*3*10^-3*(v/0.05)^2 = 0.6*9.8*0.9

===> v = 1.64 m/s <<<<<<<---------------Answer