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Question involving moment of inertia, rotation about a horizontal axis.

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A uniform cylinder, of radius 2a and moment of inertia $2Ma^2$ is free to rotate about its horizontal axis. A light, inextenzible string is wound round the cylinder and a particle of mass m is suspended on its free end. If the system is released from rest, find the acceleration of the particle.

    2. Relevant equations

    Torque(C)=Moment of inertia(I) x (angular accleration)$\alpha$.

    3. The attempt at a solution

    Resolving the tension in the string, we get T=mg-2ma(alpha)
    Which gives upon calculating the angular acceleration (given that I=2Ma^2)
    alpha=(mg)/(2am+Ma) which is somehow incorrect.

    Can someone correct my error?
  2. jcsd
  3. Aug 11, 2011 #2


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    Homework Helper
    Gold Member

    You need to draw two separate free body diagrams, one for the hanging mass and one for the cylinder. Then you need to write two F = ma equations and one τ = Iα equation based on these two FBDs.
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