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“Dancing” t-handle in zero g – Watch a video of a T-handle as it rotates about its intermediate axis, causing the rotation to be unstable. (An object that spins about its longest or shortest axis is stable, rotation along the intermediate axis is unstable.)

Dorothy Hamill scratch spin – Watch as Dorothy Hamill starts spinning slowly and speeds up when she decreases her angular inertia. This demonstrates the conservation of angular momentum.

PhET Balancing Act Simulation

Try to keep a lever in balance by placing items of different masses at different locations away from the fulcrum. This simulation is featured in the rotational motion lab in PHYSI-1100.

PhET Gravity and Orbits Simulation

See what happens to centripetal force and tangential velocity vectors as you play around with the masses of a star and its planet.

Video Transcript

Hello there! Welcome to lecture 8: rotational motion!

Rotational motion is all about things that, well, rotate! So far, our discussion has been focused on translational motion: motion along a straight line. But translational motion can’t help us to understand things like bicycle wheels, bowling balls, figure skaters, propellers, or even the motion of stars and planets. We need new concepts to help us understand rotational motion.

Each of the following concepts will be discussed in this video: rotational speed, tangential velocity, rotational acceleration, rotational inertia, torque, center of mass, stability, angular momentum, centripetal force, and centripetal acceleration.

Rotational speed

Rotational speed tells us how quickly (or slowly) something rotates about an axis. The symbol for rotational speed is the lowercase Greek letter omega, which looks like a squiggly w. There are many different units that we could use for rotational speed, depending on how we want to measure a rotation.

One way we can measure rotational speed is by determining how many rotations, or revolutions, an object makes in an interval of time. 

Instead, we can use degrees as a measure of rotational motion. Rotational speed would therefore measure how many degrees an object rotates in an interval of time. There are 360 degrees in one full rotation.

Finally, we can use radians as a measure of rotational motion. Rotational speed would therefore measure how many radians an object rotates in an interval of time. There are 2 pi radians in one full rotation.

Therefore, units for rotational speed can include: revolutions per minute, degrees per second, or radians per second.

Let’s take a look at an example.

A geostationary satellite orbits the Earth at a distance such that it appears not to move with respect to objects on the ground. An observer standing directly underneath the satellite would not be able to see any motion of that satellite.

We can calculate the rotational speed of a geostationary satellite, knowing that it makes a complete orbit of the Earth every 24 hours. Because 24 hours is equal to 1,440 minutes, we can say that a geostationary satellite has a rotational speed of 1 revolution divided by 1,440 minutes, or 6.94 times 10 to the -4 revolutions per minute.

In terms of degrees per second, we know that a geostationary satellite makes 360 degrees (one orbit) every 86,440 seconds. The rotational speed would therefore be 0.004 degrees per second.

If we wanted to use radians per second, we can take 2 pi and divide by 86,440 seconds, coming up with a rotational speed of 7.27 times 10 to the -5 radians per second. 

Tangential velocity

Have you ever ridden on a merry-go-round or a carousel? The way we experience speed on a rotating platform has to do with not only the rotational speed of the ride, but also on how far we are from the center of rotation. This quantity is called tangential velocity. While rotational speed is constant everywhere along a rotating object, the tangential velocity is not. 

Tangential velocity is pretty much just like the velocity we’ve discussed in previous lectures. The symbol is a lowercase v, and I may use a subscript of T to make it clear that it’s a tangential velocity. The units that we use are m/s.

To calculate tangential velocity, we need to determine the actual distance that an object moves as it travels along the spinning object per unit of time.

The distance around a circle is known as the circumference. It is equal to 2*pi*r, where r is the radius of the circle.

Therefore, the magnitude of the tangential velocity will be equal to 2 x pi x the distance it is from the center of rotation, divided by how many seconds it takes to complete one full rotation.

The closer an object is to the center, the smaller the tangential speed. The farther away it is, the higher the tangential speed.

The direction of the tangential velocity is tangent to the motion of the rotation: hence the name! If the word tangent is too scary, consider where an object ejected from a slingshot would go. That indicates the direction of the tangential velocity.

Therefore, the reason that we feel like we’re moving faster when we sit in the outer part of a carousel or merry-go-round is because our tangential velocity is higher!

In this video of a rotating bicycle wheel, pieces of tape have been placed at two different distances away from the center of rotation. The piece of white tape is 26.5 cm away from the center, and the piece of purple tape is 9 cm away from the center.

It takes 3.07 seconds for the bike wheel to spin one complete revolution. Both of the pieces of tape travel 360 degrees, or 2 pi radians, in 3.07 seconds. They therefore share equal rotational speeds, which can be expressed in RPM, degrees per second, or radians per second.

The white tape travels a farther liner distance in one revolution than the purple tape. The circumference describes this linear path. The white tape travels a distance of 1.67 meters in one revolution. The purple tape travels a shorter linear distance in one revolution, which is equal to 0.57 meters.

We can therefore calculate the tangential speeds of each piece of tape. The white tape has a tangential speed of 0.54 m/s, and the purple tape has a tangential speed of 0.18 m/s.

As the bicycle wheel rotates, the direction of the tangential velocity constantly changes.

Rotational acceleration

Rotational acceleration is not a concept we’re going to focus on in this class, but I want to bring it up briefly. Acceleration, as we know, means one of three things: (1) speeding up, (2) slowing down, and (3) changing direction. It’s possible for the rotational speed of an object to change, so we need to identify that rotational acceleration does exist. 

Just as we saw that force generates changes in linear motion, and that inertia resists changes in linear motion: we’ll see that torque generates changes in rotational motion, and that rotational inertia resists changes in rotational motion. That’s all!

Rotational inertia

Rotational inertia is a measure of how much an object resists changes to its rotational speed. If an object is not rotating, rotational inertia is related to how difficult it is to get that object to start rotating. If it’s already rotating, rotational inertia is related to how hard it is to make that rotation speed up or slow down.

Rotational inertia comes from two things: (1) the mass of an object, and (2) how that mass is distributed with relation to the center of rotation of the object. The more massive an object is: the harder it is to get it to start rotating. The distribution of that mass is really important as well!

I’m going to cause a hammer to rotate. If you have a hammer at home, you may want to try this out. Be careful not to move anything too fast so you don’t cause damage or injury to anything. 

First, I’m going to hold the hammer by its head. Most of the mass is now really concentrated in my hand, which is the center of rotation. It’s quite easy for me to make the hammer rotate around. Let’s see how many times I can rotate it back and forth in 5 seconds.

Next, I’m going to hold the hammer by the end of the handle. Most of the mass is now really far away from the center of rotation. Once again I’m going to count how many times I can rotate it back and forth in 5 seconds. This is definitely much harder to do than the other way around, due to its increased rotational inertia.

The farther away the mass is from the center of rotation, the higher the rotational inertia!

Let’s see one more example of this.

In this demo, there are two objects that are very similar to each other: a hoop and a disk. The hoop is hollow on the inside, and the disk is solid. But they have the same radius and the same mass.

Based on our last experiment with the hammer, we can predict what will happen when we try to make these objects rotate: the hoop is going to have a larger rotational inertia because its mass is far away from its center. So it should have a hard time rotating.

We’ll test this by placing the objects on a ramp. The two objects will rotate down to the bottom of the ramp, so whichever one has the highest rotational inertia should take longer to rotate down.

It looks like our predictions were verified! The hoop took longer to get to the end of the ramp. That means it was harder to make it rotate, and it had a higher rotational inertia than the disk. 

We’re not going to use any equations to quantify rotational inertia in this class. I will only expect you to understand this concept qualitatively.


Torque is a twist or a turn that causes an object to change its rotational motion. Just as force creates acceleration and makes something moving in a straight line speed up, slow down, or change direction… torque creates rotational acceleration and causes the rotation to speed up, slow down, or change direction. Torque has a symbol of the lowercase Greek letter tau. The units are Newton-meters. (Note: in this case, because torque is so different from concepts such as work and energy, we do not simplify this as Joules, but just use Newton-meters.)

To understand how torque works, let’s start with a lever. This is just a meter stick that’s attached to a fulcrum. It’s free to rotate, but for now it’s nice and balanced. It has no rotational motion.

I want to make this thing rotate. In order to do that, I need to apply a torque. I’m going to need to push or pull on this meter stick somehow to make that happen. So torque is definitely somehow related to force.

Let me try different magnitudes of force, all in the same spot. When I place 1 gram on the mass hanger, the meter stick rotates gently, indicating a small torque. 2 grams causes that rotation to occur much more quickly. 5 grams, even more so. 10 grams caused the largest change in rotation. Torque is proportional to the amount of force applied. In this demonstration, that force came from the gravitational force acting on these masses.

Second, I can try the same magnitude of force, but applied at varying distances away from the center of rotation. Placing 1 gram 15 cm away from the fulcrum causes a small amount of rotation. I can then place it 25 cm away, 35 cm away, and 45 cm away. The farther away the mass is placed (and the force is applied), the larger the rotation. Torque is proportional to the distance from the rotational axis that the force is applied.

Last, I’d like to demonstrate the effect of the angle of force on the torque. In this case, I have 100 grams of mass placed 45 cm to the left of the fulcrum. I have a mass hangar 45 cm to the right of the fulcrum. I use a force probe to apply a perfectly downward force on the meter stick, and we can see that it takes 0.98 Newtons to counteract the torque of the 100 gram mass.

When I place the force probe at an angle, it’s like applying a force at an angle to the center of rotation. It’s no longer perpendicular to that axis of rotation. Because I need 0.98 Newtons pointing downward, the resultant force is going to have to become greater in order to compensate for some of the force now pointing to the left. Only the perpendicular component of force contributes to torque. The parallel component of force doesn’t do any good. As the angle increases, the resultant force must continue to increase in order to keep the meter stick in balance.

Torque is equal to the perpendicular component of force times the distance from the center of rotation. Tau = F perpendicular times r. 

Let’s use what we just learned about torque to help us determine how to most efficiently open and close a door. We want to apply a perpendicular force as far away from the center of rotation as possible. This will maximize the force and lever arm, creating a maximum amount of torque.

I tried to apply a force close to the center of rotation, very near the door hinge. It took me a lot of effort to do this with a single finger, the door did not want to budge until I applied a large amount of force. When I moved spots and applied a force far away from the door hinge, even a small amount of force applied with one finger was able to get the door to open. Maximizing our lever arm distance will maximize our torque.

Understanding that, it makes a lot of sense that the door knob is at the opposite end of the door from the hinges!

Then we need to understand the concept of a perpendicular component of force. I tried closing the door by applying a force in line with the rotational axis. Note that the door didn’t move, that’s because there was no perpendicular component! Then I applied forces at an angle, but not completely perpendicular. The door was able to swing open and closed, but not terribly efficiently. Last, I applied a force perpendicular to the axis of rotation, getting the most efficient movement of the door.

The door knob is not only placed as far from the center of rotation as possible, maximizing that lever arm distance, but also enables us to apply a perpendicular force. That’s physics!

If you’re familiar with the term “lever arm”, it has to do with that distance between the force and the center of rotation. The more lever arm we have, the easier it is to do certain things. This is why we have extensions we can put on our socket wrenches. If something is really tight, we need to apply a lot of torque. There’s only so much force we can apply, so we can use the r in our torque equation to make up for our relative lack of strength.

Let’s calculate the torque that we applied to the meter stick by placing a 10 gram mass a distance of 15 cm away from the center of rotation. All of the force was perpendicular, so this example will not require us to calculate any vector components.

The force contributing to the torque comes from the gravitational force on the mass. We want to use units of Newtons so we’ll have to convert to kilograms first. 10 grams is 0.01 kilograms. Multiply by 10 m/s squared to get a force of 0.1 Newtons.

We know that the lever arm is equal to 0.15 meters. We multiply F perpendicular times r and get 0.0015 Newton-meters of torque.

Let’s do one example with a non-perpendicular force. Say we have a force vector with a magnitude of 5 Newtons acting 20 cm away from the center of rotation. 4 N points down and 3 N points to the right. Because we’re only interested in the perpendicular component of the force, we know that only the 4 N component is going to be contributing the torque. The 3 N component does not do anything useful in this scenario.

The torque is therefore 4 N times 0.2 m which is 0.8 Newton-meters.

Mechanical equilibrium

We learned in a previous lecture that mechanical equilibrium occurs when an object is traveling at a constant velocity and occurs when the net force on an object is equal to zero. We now have to expand this definition.

Mechanical equilibrium occurs when an object is traveling at a constant velocity, at a constant rotational speed, and occurs when the net force on an object is equal to zero, and the net torque on an object is equal to zero.

We learned that forces are positive when they go up or to the right, and negative when they go down or to the left. How do we give sign to torque? We do that by looking at the direction that it would make an object rotate. If an object would rotate counterclockwise, we call that a positive torque. A clockwise rotation is generated by a negative torque.

Center of mass

Center of mass is another important concept in rotational motion.  The center of mass is the point in an object around which it rotates. And if I could somehow pick an object up at that exact point, it would balance perfectly.

Another way to define center of mass is this: the center of mass in an object is the point where, on average, the mass is distributed. In an object that has a uniform density, that’s generally in the center of the object. 

If I have a meter stick, the density is relatively uniform, so the center of mass should be in the center of the stick. If I pick up the meter stick in just the right spot, I should be able to balance it perfectly.

In an object like the hammer I used earlier, which is made out of many different materials, it’s not going to necessarily be in the center, but it’s going to be closer to where most of the mass is concentrated.

The center of mass is somewhere over the place where my finger is balancing the hammer. It’s much closer to the head than it is to the handle.

Remember the hoop from the rotational inertia demo? Where is the center of mass of the hoop? The hoop is made out of a uniform density metal, and has equal thickness all around, so the center of mass of this hoop is in the exact center.

But there’s no metal there. That means it’s possible for the center of mass of an object to be located where there’s no actual mass! Cool huh? 

Maybe you’ve heard the term center of gravity, or CG. What’s the difference between that and the center of mass? The answer is: they’re almost the same thing. Center of mass has to do with how the mass of an object is distributed. Center of gravity has to do with how the weight of an object is distributed. 

If the force of gravity is equal at every spot in an object, then CM and CG are equal. That’s true for most objects most of the time. The only time we might see a difference is in really tall objects, like skyscrapers.

As of the time this video was filmed, the Burj Khalifa was the tallest building in the world, standing at an impressive 830 meters tall. Assuming that the Burj Khalifa is a rectangular cuboid of uniform density, which isn’t really accurate, then the center of mass would be 415 meters from top and bottom, in the exact center of the building.

Because the force of gravity changes with respect to the distance from the center of the Earth, the parts of this building that are closer to the ground experience more gravitational force than the parts of this building that are higher in the air.

Therefore, the weight distribution is going to be slightly closer to the bottom and will not be collocated with the center of mass. By how much? Again, assuming that the Burj Khalifa were a solid rectangle of uniform density, the center of gravity would be only a couple of millimeters below the center of mass. That’s not much of a difference.

So for the most part, it’s perfectly valid to say that center of mass and center of gravity are the same thing. But we should understand that there may be exceptions.


Simply put, stability has to do with how easily an object will topple over. More accurately, stability has to do with this. If I take an object that is in equilibrium, and I disturb it, how will its motion change? Will it return to its state of equilibrium? Or will that state of equilibrium be disrupted somehow?

I’m going to demonstrate stability with this “leaning tower.” At first, it sits in its leaning configuration without toppling over. I don’t know exactly where the center of mass is, but I know that it must be situated somewhere over the support base.

The top of the tower has a heavy mass situated in it. So by rotating the top of the tower I am able to change the location of the center of mass.

When I rotate the top of the tower by 180 degrees, it topples over to the left. That means that by rotating the top of the tower, I moved the center of mass of the object so that it was no longer over the support, but to the left of it.

An object will stay put if it has a center of mass over the support base. And an object is stable if the support base is wide enough to ensure that any motion that the object might make won’t upset that condition.

I can also use a soda can to demonstrate stability. When I place the can down normally, the center of mass is over the base, and it stays nice and steady.

When I drink just the right amount of liquid, I actually move the center of gravity down somewhat. Now there’s actually a second point on which I can rest the can. [demo] Isn’t that cool? I’ve drank just enough liquid that now the center of mass is located over this supporting base as well. However, this is not the most stable situation for the can to be in, as just a slight torque from the top or sides of the can will cause it to topple over and disturb its equilibrium.

Stability is a very important consideration in aircraft such as airplanes. Pilots have to be careful to distribute the weight of the passengers and baggage in an airplane in such a manner that the airplane remains stable. This is a routine consideration that pilots make in every flight. If you’ve ever flown on an airline and been asked to move your seat more forward in the airplane, or to a seat farther to the back, it’s probably because the pilot needed to adjust the balance of the airplane.

Angular momentum

Momentum is inertia in motion. Angular momentum is rotational inertia in rotational motion. Angular momentum is just equal to the rotational inertia multiplied by the rotational speed. We’re not going to do any calculations, but you need to know that equation to understand the topic.

Just as we saw that momentum is conserved, angular momentum is conserved as well, as long as we have no external torque acting on our system. That means that as long as there’s nothing to speed up or slow down a rotating object, its angular momentum will stay the same. Not necessarily the rotational speed: but the angular momentum.

I’ve gotten myself rotating on this chair, and I’m holding on to some dumbells. As long as I don’t do anything, and friction doesn’t slow me down too much, my angular momentum will keep me rotating at the same rotational speed.

Now as I extend my arms out, I increase my angular inertia. Because momentum is conserved, if I increase inertia, my speed will have to slow down to compensate, and it does.

When I pull my arms back in, I reduce my angular inertia. My speed will increase to compensate for that change, and to keep the angular momentum the same.

Cool, right? You’ve probably seen the same thing happen if you’ve ever watched figure skating.

Centripetal force

We’re going to talk now not about rotation but about motion along a circular path. When an object moves along a circular path at a constant speed, its direction is constantly changing. If the direction changes, we know that a force is acting on that object. We call that force centripetal force. The symbol for centripetal force is F with a subscript of C. The units are Newtons. And the equation is m v-squared over r. M is the mass of the object, v is the magnitude of the tangential velocity of the object, and r is the radius of the circle. Centripetal force always points toward the center of the circular path.

Centripetal force isn’t a force in and of itself, but is a force that must be supplied in some means to keep something moving in that circular path.

When you drive a car around a circular path, the centripetal force is supplied by friction between the tires and the pavement. The faster the car is going, the higher that force needs to be. That’s why we usually slow down to turn when we drive. Also, the tighter the turn we want to take, the more force that’s needed. If the roads are slippery, then it’s harder to get enough friction to move in a circular path.

Satellites travel in a circular path around the Earth. We’ll talk about satellites more in lecture 10, but for now we can appreciate that circular motion. The force that creates that circular motion is gravity.

I placed a small BB into a piece of coiled up plastic tubing. The BB travels in a circular path along the tubing. The normal force applied on the BB by the tubing supplies that centripetal force that allows the BB to travel in that path.

The BB has a mass of 0.33 grams, or 0.00033 kg, and travels with a radius of 10.25 cm, or 0.1025 meters. Its tangential speed is 0.46 m/s, as it travels along the circular path in 1.4 seconds. We can calculate the centripetal force acting on the BB, and find it to be 6.8 times 10 to the -4 Newtons.

If the BB were to be travelling faster, or with a smaller radius, the normal force supplied by the tubing would need to increase, and vice versa.

Centripetal acceleration

Newton’s second law tells us that forces create acceleration. Centripetal force causes an object to change its direction, which is a type of acceleration. We call that centripetal acceleration. 

Centripetal force is equal to mass times centripetal acceleration. If centripetal force is mv^2/r, then we can divide both sides by mass to find centripetal acceleration. Centripetal acceleration is equal to v^2/r. Just as with centripetal force, centripetal acceleration always points to the center of the circular path.

As I rotate this cup of water around in a circle, the tension in the string keeps it moving in that path. The centripetal force points toward the center of the circle. So what keeps the water in the cup? What stops it from sloshing on my head?

The answer is: inertia! The position of the water in the cup wants to stay where it initially was. So as long as I rotate this fast enough, the inertia is going to keep the water firmly planted in the cup.

How fast is fast enough? We need to make sure that the centripetal acceleration of the water is greater than the acceleration due to gravity. Let’s calculate the centripetal acceleration of the cup of water and confirm that it’s greater than 9.8 m/s squared.

The cup has a radius of 0.46 meters, and travels one rotation in 0.63 seconds, giving it a tangential speed of 4.59 m/s. Therefore, we can calculate the centripetal acceleration of the cup to be 45.75 m/s squared. If we divide by 9.8 m/s squared, we see that this centripetal acceleration is equal to 4.67 “g’s.” In other words, this cup is accelerating at 4.67 times the acceleration due to gravity.

Thanks for taking the time to learn about rotational motion! Until next time, stay well.