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Rotational Equilibrium and Newton's First Law in Rotational Form

AP Physics 1· difficulty 3/5

A uniform meter stick of mass MM is pivoted at one end and released from the horizontal position. What is the angular acceleration just after release? (Use I=13ML2I = \tfrac{1}{3}ML^2 about the end, g≈10 m/s2g \approx 10~\text{m/s}^2, L=1.0 mL = 1.0~\text{m}.)

  • A

    5 rad/s25~\text{rad/s}^2

  • B

    15 rad/s215~\text{rad/s}^2

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  • C

    30 rad/s230~\text{rad/s}^2

  • D

    10 rad/s210~\text{rad/s}^2

Explanation

Gravity acts at the center of mass, L/2L/2 from the pivot. τ=Mg(L/2)\tau = Mg(L/2) and I=13ML2I = \tfrac{1}{3}ML^2, so α=τ/I=Mg(L/2)13ML2=3g2L=3(10)2(1.0)=15 rad/s2\alpha = \tau/I = \dfrac{Mg(L/2)}{\frac{1}{3}ML^2} = \dfrac{3g}{2L} = \dfrac{3(10)}{2(1.0)} = 15~\text{rad/s}^2.

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