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Newton's Second Law in Rotational Form

AP Physics 1· difficulty 4/5

M, R m

A solid uniform disk pulley of mass MM and radius RR (I=12MR2I = \tfrac{1}{2}MR^2) has a light string wound around it. A mass mm hangs from the string and is released from rest. What is the linear acceleration of the hanging mass?

  • A

    a=mg/(m+2M)a = mg/(m + 2M)

  • B

    a=(m+M)g/(m+M/2)a = (m+M)g/(m + M/2)

  • C

    a=mg/(m+M)a = mg/(m + M)

  • D

    a=mg/(m+M/2)a = mg/(m + M/2)

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Explanation

For the mass: mg−T=mamg - T = ma. For the disk: TR=Iα=12MR2⋅(a/R)TR = I\alpha = \tfrac{1}{2}MR^2 \cdot (a/R), so T=Ma/2T = Ma/2. Combining gives a=mg/(m+M/2)a = mg/(m + M/2).

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