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.

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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