Video Transcript
In this video, we’re going to learn
about work and power for rotational motion. We’ll see how these two terms are
defined in a rotational scenario and how to work with them practically.
To get started, imagine that, one
day, tired of mixing chocolate chip cookie batches by hand and unable to find an
electrically powered mixer nearby, you decide to rig up your bicycle to power a
handheld mixer to make the mixing easier. You’re interested to figure out
just how much work it takes to mix a batch of cookies using this setup.
To find out, it will be helpful to
know about work and power for rotational motion. When it comes to work, we know
that, from a translational perspective, the work a given force does on an object is
equal to the magnitude of that force multiplied by the displacement of the object
𝑑. When we consider moving this
expression from a linear perspective to a rotational perspective, we can rely on the
correspondences between linear and rotational variables.
For example, for the linear
variable force, we know there’s a corresponding rotational variable, torque. And for the linear variable
distance traveled, there’s a rotational variable 𝜃, the angular distance
traveled. These correspondences suggest that
if we were to write a rotational version of the work done on an object, it would be
equal to the torque on that object multiplied by the angle through which the object
moves. And that’s correct. This is the expression of work done
from a rotational aspect.
As an example of rotational work,
this might involve something as simple as rolling a spare tire along the ground by
applying a force to the edge of the tire with your hand. That force applied some distance
away from the tire’s axis of rotation will cause the tire to rotate through some
angular distance we could call 𝜃. By multiplying the force you apply
times 𝑅, we find the torque. And the product of that torque with
the angular distance through which the tire moves is the overall rotational work
done on the tire by your hand.
Considering the movement of this
tire further, we can also say that the rotational work done on the tire is equal to
its change in rotational kinetic energy. The theorem that’s known as the
work energy theorem makes this claim. That the rotational work done on an
object is equal to the change in that object’s rotational kinetic energy.
Recalling that the rotational
kinetic energy of an object is equal to one-half its moment of inertia times its
angular speed 𝜔 squared, we can expand on our statement of the work-energy theorem
to write that 𝜏, the torque, times the angle through which an object moves is equal
to one-half its moment of inertia times its change in angular speed squared. This way of writing the work-energy
theorem assumes that the moment of inertia is constant throughout the interaction,
which is often the case.
Now that we know a bit about
rotational work, let’s consider power from a rotational perspective. We know that, in general, power is
equal to the time rate of change of work. So if we wanted to solve for
rotational power, we would take the time derivative of rotational work done. When we consider taking the time
derivative of torque times the angular displacement 𝜃, often what we find is that
torque is constant through an interaction. While in order for there to be
rotation going on, 𝜃 must change in time. This means that often we can bring
the torque, 𝜏, outside the time derivative sign so that power from a rotational
perspective equals torque times the time rate of change of 𝜃.
This term 𝑑𝜃 𝑑𝑡 is equal to
another rotational variable, which is the angular speed 𝜔. So the rotational power acting on
an object is equal to the torque on that object multiplied by its angular speed. And we’ve seen that this expression
connects with our expression for rotational work done. It’s the time derivative of
rotational work. Let’s get some practice with these
concepts of rotational work and power through an example.
A solid sphere of mass 48 kilograms
is rolling in a straight line at a speed of 6.0 meters per second across a
horizontal surface. What is the magnitude of work
required to bring the sphere to rest?
We’ll label this work magnitude
capital 𝑊. And we’ll rely on the work-energy
theorem to help us solve for it. This theorem tells us that the work
done on an object is equal to that object’s change in kinetic energy. So in our case, if we can calculate
the change in this solid sphere’s kinetic energy as it goes from moving at 6.0
meters per second to a stop, then we’ll have solved for the work done on it.
If we make a sketch of this solid
sphere as it rolls along flat ground, we know that not only is it moving linearly or
translationally with speed 𝑣, but it must also be rotating about its own
center. The total kinetic energy of this
rolling sphere will involve adding its rotational kinetic energy along to its linear
or translational kinetic energy.
In order to solve then for the
sphere’s overall change in kinetic energy, we’ll recall the relationships for
rotational and translational kinetic energy. Rotational kinetic energy is equal
to an object’s moment of inertia, 𝐼, times its angular speed squared. And translational kinetic energy is
equal to one-half its mass times its linear speed squared. Writing these into our equation, we
know because we’ve been given 𝑚 and 𝑣, we can solve for those values. But we don’t yet know 𝐼 or 𝜔.
If we consider first the moment of
inertia 𝐼, we know that because we’re working with a solid sphere and it’s rotating
about its center, when we look up the moment of inertia of such a shape rotating in
such a way, we find it’s equal to two-fifths the mass of the sphere times its radius
squared. Plugging that expression in for 𝐼,
now we focus on the angular speed 𝜔.
We’ll recall that there’s a
relationship between linear speed 𝑣 and angular speed 𝜔. They’re related by the radius 𝑟,
𝑣 equals 𝑟𝜔. This means we can replace 𝜔 in our
expression with 𝑣 over 𝑟. And we see that when we do, the
factors of 𝑟 squared in this term cancel out, which is good because we don’t know
the radius of the sphere.
Our equation for Δ𝐾𝐸 then
simplifies to one-fifth 𝑚𝑣 squared plus one-half 𝑚𝑣 squared. This is equal to seven-tenths 𝑚𝑣
squared or plugging in for 𝑚 and 𝑣 seven-tenths times 48 kilograms times 6.0
meters per second squared. When we multiply out this
expression, to two significant figures, we find a result of 1.2 kilojoules. That’s the change in kinetic energy
of the solid sphere as it comes to a stop. And by the work energy theorem,
that’s also the work done on the sphere to stop it.
Let’s summarize what we’ve learned
so far about work and power for rotational motion. We’ve seen that the rotational work
done on an object is the torque applied times the angular distance through which the
object moves. As an equation, rotational work is
equal to 𝜏 times 𝜃. We’ve also seen that rotational
power is the time rate of change of work. Rotational power equals 𝑑𝑊 sub 𝑟
𝑑𝑡, which equals 𝜏 times 𝜔, 𝜔 being the time rate of change of the angular
displacement 𝜃. And finally, we’ve seen that the
work-energy theorem connects rotational work with rotational kinetic energy. It says that the rotational work
done on an object is equal to that object’s change in rotational kinetic energy.
