Video Transcript
In this video, we’re going to learn
about rotational kinetic energy. We’ll see what this term means and
how it relates to translational kinetic energy.
To start off, say that you put two
shapes at the top of an inclined plane: one is a perfect sphere and the other is a
perfect cube. If both of these objects have the
exact same mass and then released to slide down this plane which we can assume is
completely frictionless, then when they get to the bottom of the plane, which shape
we wonder will have the most kinetic energy.
Rotational kinetic energy is the
energy that an object has simply because it’s rotating. When we first hear this term
“rotational kinetic energy,” we might naturally think of the kinetic energy we’ve
learned of before which has to do with the translation. When a massive object is in motion
— that is it has some speed — then we can calculate its translational kinetic energy
with this equation. But translation that is moving in a
line is only one type of motion. Rotation is another type.
When it comes to connecting
rotation and translation — these two types of motion — we’ve seen that variables in
one type have corresponding values in the other type. For example, linear speed 𝑣
corresponds to angular speed 𝜔 and linear mass 𝑚 corresponds to rotational mass
moment of inertia. So if we had to guess the
mathematical relationship telling us what an object’s rotational kinetic energy is,
we might base it on these correspondences and also the form of the translational
kinetic energy we’re more familiar with.
Our educated guess might be that an
object’s rotational kinetic energy is one-half its moment of inertia multiplied by
its angular speed squared. It turns out this is correct. It’s an accurate representation of
the kinetic energy an object has because it’s turning or rotating. To get a sense for why this is the
case, let’s imagine we have a circular disc that’s rotating about its centre. It’s not translating at all. But it is rotating and we want to
solve for the kinetic energy of this rotating object.
If we didn’t know the rotational
kinetic energy equation, if we had never seen it before and weren’t familiar with
it, we would probably start with the equation we’ve seen before for translational
kinetic energy. And since this expression has to do
with an object’s mass and its speed, what we might start to do is looking at this
rotating disk, we might start to pick out tiny little chunks of the disc and call
those individual masses.
So we might say that this piece
over here is 𝑚 one and the tiny chunk over in this part of the disc we’ll call 𝑚
two and so forth and so on until we’ve broken the disc up into a bunch of very tiny
masses. We know that because the disc is
rotating, each one of the masses we’ve picked out is also in motion. And depending on a mass points
distance to the axis of rotation, that linear speed will be more or less.
Now that the disc is broken up into
lots of tiny masses, each of which has its own speed, we can solve for the kinetic
energy of the disc overall by adding up the kinetic energies of each of these
individual masses. We could write that this disc’s
total kinetic energy is equal to the sum over the index 𝑖 of all the individual
masses of the disc multiplied by the speeds of those individual masses squared
multiplied by a half.
Looking at this linear speed term
𝑣 sub 𝑖, we can recall that when your speed in general is equal to 𝑟 the radius
of whatever circular object we’re speaking of multiplied by the angular speed 𝜔, we
can substitute in then 𝑟 sub 𝑖 times 𝜔 for 𝑣 sub 𝑖. And the reason that 𝜔 doesn’t have
an 𝑖 subscript is because it’s the same for every mass element all throughout this
rotating disc. And since it’s the same, we can
factor out the 𝜔 squared term along with the one-half. So now, our overall kinetic energy
of our rotating disc is 𝜔 squared over two multiplied by the sum of the individual
mass elements times the radius of each mass element from the centre of the disc
squared.
Earlier, we brought up moment of
inertia as a rotational version of mass. And we may recall that the moment
of inertia of a point mass that is an infinitely small chunk of matter is equal to
the mass of that chunk times the distance it is from its axis of rotation
squared. Looking at this form for 𝐼 sub
𝑝𝑚, we see that it matches what’s left in our summation — 𝑚 sub 𝑖 times 𝑟 sub
𝑖 squared. So we can replace 𝑚 sub 𝑖 𝑟 sub
𝑖 with 𝐼 sub 𝑖 the moment of inertia of a point mass summed over all the point
masses of this disc.
Another way to write this is that
the kinetic energy is one-half the sum of all the moments of inertia of our
individual mass elements multiplied by the angular speed of our disc squared. This is the rotational kinetic
energy of our disc and we found it by applying the translational kinetic energy
relationship. Looking at this expression, if we
had an object, whose moment of inertia 𝐼, we could solve for that included the
whole object, then from that we see how we would get the general expression for
rotational kinetic energy of one-half 𝐼 times 𝜔 squared.
One last thing about rotational
kinetic energy and that’s that an object can have rotational as well as
translational kinetic energy as in the case of our rolling ball down the incline at
the start of the segment. In cases like that, the total
kinetic energy of our object is equal to the translational kinetic energy plus the
rotational. The object has both and so we’ll
count for both.
Let’s get some practice now with an
example that connects rotational and translational kinetic energy.
A small helicopter is propelled
horizontally at a speed of 20.0 metres per second by the rotation at 300 rpm of four
rotor blades. Each blade is 50.0 kilograms in
mass, 4.00 metres in length, and can be modelled as a thin rod that is joined at one
end to an axis of rotation that is perpendicular to its length. The mass of the helicopter
including its rotor blades is 1000 kilograms. Find the rotational kinetic energy
of the blades. Find the ratio of the rotational
kinetic energy of the blades to the translational kinetic energy of the
helicopter.
In this two-part exercise, we first
want to solve for the rotational kinetic energy of the blades of the helicopter and
then we want to solve for the ratio of that rotational kinetic energy to the
translational kinetic energy of the helicopter overall. Let’s start out with a diagram of
the helicopter in flight.
If we consider a top down view of
the helicopter while it’s flying, we know that its main rotor is spinning with an
angular speed — we’ve called capital Ω — given as 300 revolutions per minute. And while this is happening, the
helicopter is translating along forward at a linear speed — we’ve called 𝑣 — given
as 20.0 metres per second. We’re also told that each of the
four blades in the main rotor has a length of 4.00 metres and a mass of 50.0
kilograms.
Knowing all this, we want to
calculate the rotational kinetic energy of the blades in the rotor. We can recall that an object’s
rotational kinetic energy is equal to one-half its moment of inertia times its
angular speed in radians per second squared. The statement tells us that each
one of the blades in the main rotor can be modelled as a thin rod rotating
perpendicularly about one of its ends.
When we look up the moment of
inertia of an object like this rotating about an axis like this, we find that it’s
equal to one-third the object’s mass times its length squared. This all means that the rotational
kinetic energy of one of the four rotor blades is equal to one-half times a third
times the mass of the blade times its length squared times its angular speed 𝜔
squared.
We aren’t given that angular speed
𝜔 in units of radians per second. But we are given that speed in
units of revolutions per minute. To convert 300 revolutions per
minute into units of radians per second, we can first multiply this fraction by one
minute every 60 seconds, which will cancel out the units of minutes and give us a
time unit of seconds. Then since there are two 𝜋 radians
in every one complete circular revolution, we multiply by this ratio which leads us
to cancelling out the units for revolutions.
We find then that the angular speed
of these rotor blades in units of radians per second is 300 times two 𝜋 over
60. We now have an expression for the
rotational kinetic energy of one of our rotor blades in terms of values that are all
known. In order to find the total
rotational kinetic energy of the entire main rotor, we multiply this term by four
which creates a leading factor of two in place of our one-half previously.
We’re now ready to plug in for 𝑚,
𝐿, and 𝜔 to solve for KE sub 𝑟. When we do and then enter this
expression on our calculator, we find that to three significant figures KE sub 𝑟 is
5.26 times 10 to the fifth joules. That’s the rotational kinetic
energy of the blades in the main rotor of this helicopter.
Next, we want to solve for the
ratio of the rotational to translational kinetic energy of this helicopter. Recalling that translational
kinetic energy is one-half an object’s mass times its speed squared, we can say that
the translational kinetic energy of this helicopter as it flies is one-half the mass
of the helicopter 𝑚 sub ℎ times its speed 𝑣 squared.
In the problem statement, we’re
told that the overall mass of the helicopter 𝑚 sub ℎ is 1000 kilograms. Since we already know 𝑣, the
linear speed of the helicopter, we’re ready to plug in and solve for KE sub 𝑡. When we plug in for 𝑚 sub ℎ and 𝑣
and enter this expression on our calculator, we find a result for the overall
translational kinetic energy of the helicopter. We divide this result into the
rotational kinetic energy of the blades in the helicopter’s rotor to solve for the
ratio we’re interested in. To three significant figures, this
ratio is 2.63. So the blades of the main rotor of
the helicopter have 2.63 times as much kinetic energy rotationally as the helicopter
has overall translationally.
Let’s summarize what we’ve learned
about rotational kinetic energy. We’ve seen that rotational kinetic
energy is the energy an object has because it’s rotating. Written as an equation, we’ve seen
that rotational kinetic energy is equal to one-half an object’s moment of inertia
times its angular speed in radians per second squared. We’ve also seen that the rotational
kinetic energy of an object is found by adding up all the translational kinetic
energies of its mass elements. And finally, we saw that if an
object rotates as well as translates, for example, a ball rolling down an incline,
then its total kinetic energy is equal to the sum of its rotational and its
translational kinetic energy.
