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Newton's Second Law in Rotational Form

AP Physics 1· difficulty 3/5

A solid disk pulley of mass M=2.0 kgM = 2.0~\text{kg} and radius R=0.10 mR = 0.10~\text{m} is free to rotate on its axle. A string wound around it holds a 1.0 kg1.0~\text{kg} mass. Using g≈10 m/s2g \approx 10~\text{m/s}^2, what is the hanging mass's downward acceleration? (Disk: I=12MR2I = \tfrac{1}{2}MR^2.)

  • A

    10 m/s210~\text{m/s}^2

  • B

    5.0 m/s25.0~\text{m/s}^2

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  • C

    2.5 m/s22.5~\text{m/s}^2

  • D

    6.7 m/s26.7~\text{m/s}^2

Explanation

Equations: mg−T=mamg - T = ma and TR=Iα=12MR2(a/R)TR = I\alpha = \tfrac{1}{2}MR^2 (a/R), so T=12MaT = \tfrac{1}{2}M a. Substituting: mg=a(m+M/2)⇒a=g m/(m+M/2)=10(1)/(1+1)=5.0 m/s2mg = a(m + M/2) \Rightarrow a = g\,m/(m + M/2) = 10(1)/(1 + 1) = 5.0~\text{m/s}^2.

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