A 5.0-kg hollow cylinder of radius 0.16m rotates freely about an axle that runs

Q: A 5.0-kg hollow cylinder of radius 0.16m rotates freely about an axle that runs

torque = F*R = 50*0.16 = 8 Nm Work = Torque*theta = 8*1000 = 8000 J part B) I = m*R^2 = 5*0.16*0.16 = 0.128 kg m^2 Work = 0.5*I*(w2^2 - w1^2) W = 0.5*I*w2^2 w2 = sqrt((2*W)/I) w2 = 353.55 rad/s

ID: 1293084 • Letter: A

Question

A 5.0-kg hollow cylinder of radius 0.16m rotates freely about an axle that runs through its center and along its long axis. A cord is wrapped around the cylinder and is pulled straight from the cylinder with a steady tensile force of 50 N . As the cord unwinds, the cylinder rotates, with no slippage between cord and cylinder.

Part A

Calculate the work done by the tensile force as the cylinder rotates through 1000 rad .

Express your answer with the appropriate units.

Part B

If the cylinder starts from rest, calculate its rotational speed after it has rotated through 1000 rad .

Explanation / Answer

torque = F*R = 50*0.16 = 8 Nm

Work = Torque*theta = 8*1000 = 8000 J

part B)

I = m*R^2 = 5*0.16*0.16 = 0.128 kg m^2

Work = 0.5*I*(w2^2 - w1^2)

W = 0.5*I*w2^2

w2 = sqrt((2*W)/I)

w2 = 353.55 rad/s

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A 5.0-kg hollow cylinder of radius 0.16m rotates freely about an axle that runs

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