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5 February, 07:33

A thin, horizontal rod with length l and mass M pivots about a vertical axis at one end. A force with constant magnitude F is applied to the other end, causing the rod to rotate in a horizontal plane. The force is maintained perpendicular to the rod and to the axis of rotation.

Calculate the magnitude of the angular acceleration of the rod.

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  1. 5 February, 07:42
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    The magnitude of the angular acceleration is α = (3 * F) / (M * L)

    Explanation:

    using the equation of torque to the bar on the pivot, we have to:

    τ = I * α, where

    I = moment of inertia

    α = angular acceleration

    τ = torque

    The moment of inertia is equal to:

    I = (M * L^2) / 3

    Also torque is equal to:

    τ = F * L

    Replacing:

    I * α = F * L

    α = (F * L) / I = (F * L) / ((M * L^2) / 3) = (3 * F) / (M * L)
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