(hrwc10p67) A uniform spherical shell of mass M = 6.5 kg and radius R = 18.0 cm

Question

(hrwc10p67) A uniform spherical shell of mass M = 6.5 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 = 2.75×10-3 kg m2 and radius r = 5.0 cm, and its attached to a small object of mass m = 2.0 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

Moment of inertia of hallow sphere, I1 = (2/3)*M*r^2

= (2/3)*6.5*0.18^2

= 0.1404 kg.m^2

moment of inertia of pulley, I2 = 2.75*10^-3 kg.m^2

Let v is the speed of hanging object when it fallen 2m,

anglar speed of sphere, w1 = v/R

= v/0.18

angular speed of pulley, w2 = v/r

= v/0.05

now Apply conservation of energy

loss of potentail energy = gain in kinetic energy

m*g*h = 0.5*m*v^2 + 0.5*I1*w1^2 + 0.5*I2*w2^2

2*9.8*2 = 0.5*2*v^2 + 0.5*0.1404*(v/0.18)^2 + 0.5*2.75*10^-3*(v/0.05)^2

39.2 = v^2 + 17.6*v^2 + 0.55*v^2

39.2 = 19.15*v^2

==> v = sqrt(39.2/19.15)

= 1.43 m/s <<<<<<----------Answer

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