What is the angular acceleration?

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.


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=mg-T (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:


    (mg-ma)r=(1/2)mr^2α (m and one r get canceled from both sides)



    (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)

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