A thin, light string is wrapped around the rim of a 4.15 kg solid uniform disk t

Question

A thin, light string is wrapped around the rim of a 4.15 kg solid uniform disk that is 28.5 cm in diameter. A person pulls on the string with a constant force of 103.0 N 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.

A:

Find the angular acceleration of the disk about its center of mass.

B:Find the linear acceleration of its center of mass
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?
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

Torque T = I*alpha

I is the moment of inertia = 0.5*M*R^2 = 0.5*4.15*(0.285/2)^2 = 0.042135 kg*m^2

alpha = angular accelaration

but torque T = r*F = (0.285/2)*103 = 14.6775 N-m

then T = I*alpha

alpha = T/I = 14.6775/0.042135 = 348.4 rad/s^2

B) linear accelaration a = r*alpha = (0.285/2)*348.4 = 49.647 m/s^2

C) moment of inertia of disk and cylinder are same,hence the angulae accelaration alpha is same

alpha_cylinder = 384.4 rad/s^2

D) linear accelaration a = 49.647 m/s^2 is also same

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