A uniform spherical shell of mass M = 1.0 kg and radius R = 20.0 cm rotates abou

Question

A uniform spherical shell of mass M = 1.0 kg and radius R = 20.0 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 = 2.70×10-3kg 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 = 1.0 m from rest: Use work - energy considerations.

Explanation / Answer

The pulley and the sphere acquire rotational kinetic energy.

mgh = 0.5mv2 + 0.5IpWp2 + 0.5IsWs2

where "Ip" is the rotational inertia of the pulley,

"Wp" is the angular velocity of the pulley,

"Is" is the rotational inertia of the sphere and

"Ws" is the angular velocity of the Sphere.

Wp = V/Rp

Ws = V/Rs

mgh = 0.5mv^2 + 0.5IpWp2 + 0.5IsWs^2

The rotational inertia of a spherical shell is: (2/3)M(R^2

For the Sphere Shell, Is = (2*1* 0.2* 0.2) = 0.08 kg*m^2

thus mgh = 0.5V^2(m + Ip/Rp^2 + Is/Rs^2 )

V=sqrt(2mgh/(m + Ip/Rp^2 + Is/Rs^2))

=sqrt(2* 1* 9.8* 1/(1+0.75 + 2)

V = 2.286 m/s

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