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2 years ago

12

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 ap

plied 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.

1 answer:

katen-ka-za [31]2 years ago

8 0

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)

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