A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter .4m and hangs ver?

A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter .4m and hangs vertically, as shown. The rotational inertia of the wheel is .5kg.m^2. When the block is released and the cord unwinds, find the acceleration of the block?

Could you please help me with this problem?

Thank you

1 Answer

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  • Ivan A
    Lv 6
    9 years ago
    Favorite Answer

    Free-body analysis of the forces on the 16 Kg block gives

    mg - T = ma

    and a torque analysis around the flywheel

    T R = I α

    From the kinematics of the system we have that

    α R = a

    substituting into the expression for torque and solving for T we get

    T = I a / R^2

    substituting into the forces of the free body diagram

    mg - Ia/R^2 = ma

    solving for a

    a = mg / (m + I/R^2) = 16*9.8/(16+0.5/0.4^2) = 8.19 m/s^2

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