Rolling

AP Physics 1· difficulty 4/5

θ I = (2/5)MR²

A solid uniform sphere (I=25MR2I = \tfrac{2}{5}MR^2) rolls without slipping from rest down an incline of angle θ\theta. What is the linear acceleration of the center of mass along the incline?

  • A

    a=gsinθa = g\sin\theta

  • B

    a=(1/2)gsinθa = (1/2)g\sin\theta

  • C

    a=(5/7)gsinθa = (5/7)g\sin\theta

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  • D

    a=(2/3)gsinθa = (2/3)g\sin\theta

Explanation

With rolling, energy conservation gives 12mv2(1+I/(mR2))=mgh\tfrac{1}{2}m v^2(1 + I/(mR^2)) = mgh. For solid sphere I/(mR2)=2/5I/(mR^2) = 2/5, so effective mass factor is 7/57/5 and a=gsinθ/(1+2/5)=(5/7)gsinθa = g\sin\theta/(1 + 2/5) = (5/7)g\sin\theta.

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