Video Transcript
A uniform rod of length two 𝑙 and
mass 𝑚 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 24𝑔 over 𝑙,
where 𝑔 is the acceleration due to gravity. Find the magnitude of the force
applied.
Let’s start with a diagram of the
situation. We have a uniform rod of length two
𝑙 and mass 𝑚. A uniform rod refers to
one-dimensional object with length but no width. But let’s draw it with some width
just to make things a little clearer. So this rod has length two 𝑙 and
mass 𝑚. And it’s rotating about a vertical
axis through its center with an angular acceleration of 24𝑔 over 𝑙. And we have a force 𝐅 applies to
the end of the rod, which causes this angular acceleration. This is the quantity we wish to
find.
We begin with Newton’s second law
of motion in its rotational form. The sum of all the torques Τ on a
body is equal to the moment of inertia of that body 𝐼 times the angular
acceleration 𝜔 prime. We need to find an expression for
𝐅 from this equation. A torque Τ is given by the product
of the perpendicular component of the force 𝐅 times its distance from the axis of
rotation 𝑑.
In our case, the force applied 𝐅
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 𝑙, the distance 𝑑 is just equal
to 𝑙. So the force 𝐅 is given by Τ over
𝑙.
Substituting in the expression for
Τ from Newton’s second law gives us 𝐼𝜔 prime over 𝑙. We already know that 𝜔 prime, the
angular acceleration, is 24𝑔 over 𝑙. So our expression for 𝐅 becomes
24𝑔𝐼 over 𝑙 squared. We now have only one unknown in our
expression for 𝐅, the moment of inertia of the rod 𝐼.
For a one-dimensional object such
as this rod, the moment of inertia 𝐼 is given by the integral between the two
endpoints 𝑥 one and 𝑥 two of the square of the distance from the axis of rotation
𝑥 with respect to the mass element d𝑚.
Going back to our diagram of the
rod, consider a small piece of the rod, shown here in orange. It has mass d𝑚 and length d𝑥. We want to integrate with respect
to the length d𝑥. So we need to find an expression
for d𝑚 in terms of d𝑥.
Since density is mass divided by
volume, or in this case linear density is mass divided by length, then the linear
density 𝜌 is equal to d𝑚 over d𝑥. Therefore, d𝑚 equals 𝜌 times
d𝑥. 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 𝜌 is equal
to 𝑚 over two 𝑙. Therefore, d𝑚 is equal to 𝑚d𝑥
over two 𝑙.
Therefore, the moment of inertia 𝐼
is given by the integral between 𝑥 one and 𝑥 two of 𝑥 squared times 𝑚 over two
𝑙 d𝑥. 𝑚 over two 𝑙 is just a constant,
so we can move it outside the integration. The limits 𝑥 one and 𝑥 two are
just the start and endpoints of the rod with respect to the axis of rotation. The axis of rotation has an
𝑥-value of zero. So the start point has an 𝑥-value
of negative 𝑙 and the endpoint has a value of 𝑙. So we have 𝑚 over two 𝑙 times the
integral between negative 𝑙 and 𝑙 of 𝑥 squared d𝑥.
Integrating with respect to 𝑥
gives us one-third 𝑥 cubed. Evaluating and simplifying gives us
𝑚𝑙 squared over three. Now, substituting this back into
our equation for 𝐅, this gives us 24𝑔 over 𝑙 squared times 𝑚𝑙 squared over
three. And simplifying gives us our
answer: 𝐅 equals eight 𝑚𝑔.
