Video: Rotational Kinetic Energy

In this video we derive the rotational kinetic energy of a spinning object and calculate an object’s rotational and translational kinetic energy.

11:34

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.

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