Question

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.

Answer 1

Answer:

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)