Question

A string is wrapped around a uniform solid cylinder of radius r. The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m. Note that the positive y direction is downward and counterclockwise torques are positive.

Find the magnitude, alpha, of the angular acceleration of the cylinder as the block descends.

Express your answer in terms of the cylinder's radius r and the magnitude of the acceleration due to gravity g.

Answer 1

Let T be the string tension. That and the angular acceleration alpha are the unknowns. Separate equations for the acceleration of the block and the angular acceleration of the cylinder will allow you to solve for both variables. The moment of inertia of the cylinder in (1/2) M r^2

m g - T = m a

T*r = torque = I*alpha = (1/2) m r^2 * alpha
T = (1/2) m*r*alpha

m g = ma + (1/2) m r * alpha
= ma + (1/2) m a
(since alpha = a/r)

a = (2/3) g
alpha = (2/3)(g/r)

Answer 2

Let T be the string tension. That and the angular acceleration alpha are the unknowns. Separate equations for the acceleration of the block and the angular acceleration of the cylinder will allow you to solve for both variables. The moment of inertia of the cylinder in (1/2) M r^2

m g - T = m a

T*r = torque = I*alpha = (1/2) m r^2 * alpha
T = (1/2) m*r*alpha

m g = ma + (1/2) m r * alpha
= ma + (1/2) m a
(since alpha = a/r)

alpha = (2/3) g
Note that the magnitude of the linear acceleration of the block is 23g , which does not depend on the value of r .