### Video Transcript

A uniform rod of length two π and
mass π is free to rotate in a horizontal plane about a vertical axis passing
through its center. A horizontal force of constant
magnitude is applied to one of the ends of the rod in a direction perpendicular to
the rod, causing the rod to rotate with an angular acceleration of 24π over π,
where π is the acceleration due to gravity. Find the magnitude of the force
applied.

Letβs start with a diagram of the
situation. We have a uniform rod of length two
π and mass π. A uniform rod refers to
one-dimensional object with length but no width. But letβs draw it with some width
just to make things a little clearer. So this rod has length two π and
mass π. And itβs rotating about a vertical
axis through its center with an angular acceleration of 24π over π. And we have a force π
applies to
the end of the rod, which causes this angular acceleration. This is the quantity we wish to
find.

We begin with Newtonβs second law
of motion in its rotational form. The sum of all the torques Ξ€ on a
body is equal to the moment of inertia of that body πΌ times the angular
acceleration π prime. We need to find an expression for
π
from this equation. A torque Ξ€ is given by the product
of the perpendicular component of the force π
times its distance from the axis of
rotation π.

In our case, the force applied π
is already perpendicular to the axis of rotation. Since the axis of rotation is about
the center of the rod and the rod is of length two π, the distance π is just equal
to π. So the force π
is given by Ξ€ over
π.

Substituting in the expression for
Ξ€ from Newtonβs second law gives us πΌπ prime over π. We already know that π prime, the
angular acceleration, is 24π over π. So our expression for π
becomes
24ππΌ over π squared. We now have only one unknown in our
expression for π
, the moment of inertia of the rod πΌ.

For a one-dimensional object such
as this rod, the moment of inertia πΌ is given by the integral between the two
endpoints π₯ one and π₯ two of the square of the distance from the axis of rotation
π₯ with respect to the mass element dπ.

Going back to our diagram of the
rod, consider a small piece of the rod, shown here in orange. It has mass dπ and length dπ₯. We want to integrate with respect
to the length dπ₯. So we need to find an expression
for dπ in terms of dπ₯.

Since density is mass divided by
volume, or in this case linear density is mass divided by length, then the linear
density π is equal to dπ over dπ₯. Therefore, dπ equals π times
dπ₯. But this rod is uniform, so its
density is the same everywhere. Therefore, the density at any given
point is the same as the overall density of the rod. Therefore, the density π is equal
to π over two π. Therefore, dπ is equal to πdπ₯
over two π.

Therefore, the moment of inertia πΌ
is given by the integral between π₯ one and π₯ two of π₯ squared times π over two
π dπ₯. π over two π is just a constant,
so we can move it outside the integration. The limits π₯ one and π₯ two are
just the start and endpoints of the rod with respect to the axis of rotation. The axis of rotation has an
π₯-value of zero. So the start point has an π₯-value
of negative π and the endpoint has a value of π. So we have π over two π times the
integral between negative π and π of π₯ squared dπ₯.

Integrating with respect to π₯
gives us one-third π₯ cubed. Evaluating and simplifying gives us
ππ squared over three. Now, substituting this back into
our equation for π
, this gives us 24π over π squared times ππ squared over
three. And simplifying gives us our
answer: π
equals eight ππ.