A string is wrapped around a uniform solid cylinder of radius r, as shown in the figure (Figure 1) . 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.
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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.
1 Answer

Rule one: Always write down F=ma (usually good for partial credit)
Force on block: ma=mgT (T being tension in the string)
Moment of Inertia of the cyclinder: I=(1/2)mr^2
Torque: τ=Tr=Iα (Tensionradius=MoIangular acceleration)
Angular acceleration: α=a/r (linear acceleration/radius)
Using the torque equation we can plug in relations from the other equations:
Tr=Iα
(mgma)r=(1/2)mr^2α (m and one r get canceled from both sides)
ga=1/2rα
(ga)/r=1/2*α
(g/r)(a/r)=1/2*α (remember that a/r is equal to angular acceleration)
g/r=3/2*α (added one angular acceleration to both sides)
Therefore, angular acceleration of the cylinder is 2*gravity divided by 3*radius (2g/3r)