A uniform spherical shell of mass M = 4.5 kg and radius R = 18.0 cm rotates abou
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A uniform spherical shell of mass M = 4.5 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 = 5.40×10-3 kg m2 and radius r = 6.0 cm, and its attached to a small object of mass m = 4.0 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 = 0.9 m from rest: Use work - energy considerations.
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Rotational Kinetic Energy is KE = (1/2)*I*w^2 where "w" is the angular velocity
The tangential velocity of the sphere and the pulley are equal since they are connected by the string. The change in height of the object represents a potential energy loss, and the object acquires kinetic energy of (1/2)*m*v^2. The pulley and the sphere acquire rotational kinetic energy.
mgh = (1/2)*m*v^2 + (1/2)*Ip*Wp^2 + (1/2)*Is*Ws^2
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.
The constant is the tangential velocity of the two rotating objects, which equals the vertical velocity of the falling object. The angular velocity of a rotating object is it's tangential velocity divided by it's radius. So
Wp = V/Rp where V is the tangential velocity, and Rp is the radius of the pulley.
Ws = V/Rs where V is again tangential velocity, and Rs is the radius of the sphere
m*g*h = (1/2)*[m*V^2 + Ip*(V/Rp)^2 + Is*(V/Rs)^2]
The rotational inertia of a spherical shell is: (2/3)*M*(R^2)
For the Sphere Shell, Is = (2/3)*(4.5)/(0.18^2) = 0.0972 kg*m^2
Factor out the "V^2" from the above equation...
m*g*h = (1/2)*V^2*[m + Ip/(Rp^2) + Is/(Rs^2)]
V^2 = 2*m*g*h/[m + Ip/(Rp^2) + Is/(Rs^2)]
V = sqrt(2*m*g*h/[m + Ip/(Rp^2) + Is/(Rs^2)]
plug in the known values...
V = sqrt(70.56/[4.0 + 1.5 + 3.0]) = 2.88 m/s
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