ACK NEXT A uniform spherical shell of mass M -12.0 kg and radius R-0.420 m can r

Question

ACK NEXT A uniform spherical shell of mass M -12.0 kg and radius R-0.420 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 1-0 100 kgm2 and radius r-0.0560 m, and is attached to a small object of mass m-3.80 kg. There is 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 1.08 m after being released from rest? Use M. R the tolerance is + 2% by Study Question Attempts: 0 of 6 used SAVE FOR LATER

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 mean 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

Ip = rotational inertia of the pulley

Wp = is the angular velocity of the pulley

Is = is the rotational inertia of the sphere

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

V = tangential velocity,

Rp = radius of the pulley.

Ws = V/Rs

V = tangential velocity

Rs = 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 = (2/3)*M*(R^2)

For the Sphere Shell, Is = (2/3)*(12)/(0.42^2) =1.4112 kg*m^2

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)]

V = sqrt(80.52/[3.8+ 31.88 + 8]) = 1.35 m/s

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