Part A: Find the angular acceleration of the disk about its center of mass. (rad

Question

Part A: Find the angular acceleration of the disk about its center of mass. (rad/s^2)

Part B: Find the linear acceleration of its center of mass. (m/s^2)

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? (rad/s^2)

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? (m/s^2)

Explanation / Answer

,mass= 3.70 kg

diameter= 29.5 cm

force = 102.0 N

Angular acceleration=Torque/Moment of inertia

Moment of inertia=mr2/2= 3.7*14.752/2=402.49 kg.m2

Angular acceleration,a=102*14.75/402.49=3.73rad/s2

Linear acceleration=rXa=55.13 m/s2

For thin walled cylinder the moment of inertia=mr2

So the angular accleration and linear acceleration will reduce by half

Linear acceleration=27.56 m/s2

Angular acceleration=1.865rad/s2

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