A thin, light string is wrapped around the rim of a 3.85kg solid uniform disk th

Question

A thin, light string is wrapped around the rim of a 3.85kg solid uniform disk that is 27.5cm in diameter. A person pulls on the string with a constant force of 105.0N tangent to the disk, as shown in the figure below (Figure 1) . The disk is not attached to anything and is free to move and turn.

Part A.) Find the angular acceleration of the disk about its center of mass.

Part B.) Find the linear acceleration of its center of mass

Part C) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the angular acceleration of the disk about its center of mass?

Part D)If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the linear acceleration of its center of mass?

Explanation / Answer

A) torque = I*alfa

alfa= T/I = (F*R)/(0.5*M*R^2) = (105*0.275)/(0.5*3.85*0.275^2) = 198.3471 rad/s^

B) a = R*alfa= 54.5455 m /s^2

C) torque = I*alfa

alfa= T/I = (F*R)/(0.5*M*R^2) = (105*0.275)/(0.5*3.85*0.275^2) = 198.3471 rad/s^

D) a = R*alfa= 54.5455 m /s^2

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