A uniform spherical shell of mass M = 5.40 kg and radius R = 0.750 m can rotate
ID: 1514298 • Letter: A
Question
A uniform spherical shell of mass M = 5.40 kg and radius R = 0.750 m can rotate 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 = 0.110 kg·m2 and radius r = 0.130 m, and is attached to a small object of mass m = 4.00 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 when it has fallen a distance 0.672 m after being released from rest? Use energy considerations.
Explanation / Answer
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)*(5.40)/(.75^2) = 2.0kg*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(52.68/[4.0+ 6.5 + 9.6]) = 0.361m/s
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