Topic 5.2: Connecting Linear and Rotational Motion

Standing still

In the summer of nineteen fifty-five, at Edwards Air Force Base, on the edge of a huge dry lakebed in the Mojave Desert, test pilots started flying an Air Force airplane that never once flew faster than the speed of sound. The Air Force called it the X F eighty-four H, and people nicknamed it the Thunderscreech. It flew only twelve times, but its strangest trick happened on the ground, whenever the crew ran the engine with chocks jammed against the wheels. Every time they did, the ends of its propeller blades went faster than sound, while the airplane sat perfectly still.

Before I show you the air, I want your guess. Somewhere between the hub and the tip, the blade crosses the speed of sound, so drag the pink ring to where you think that happens, and let go.

If we could see the air the way a schlieren camera does, the trick engineers use to photograph shock waves, we’d see them peeling off the ends of the blades, the same kind a jet makes when it breaks the sound barrier, while the airplane itself doesn’t move an inch. Follow one blade out from the middle to the end and the speed keeps climbing, from almost nothing at the hub to about four hundred meters per second at the very tip, which is about one point two times the speed of sound.

By the end of this lesson, you’ll be able to take one number off a spec sheet, how fast something spins, and work out how fast any point on it is moving, and you’ll find exactly where on this propeller the sound barrier breaks. To get there, we need to slow this propeller way down and put a couple of dots of paint on one of its blades, because the whole idea is hiding in the way two dots on the same blade move.

Same turn, different trip

Here’s the same propeller slowed way down, with one dot a third of the way out, sixty-one centimeters from the center, and another right at the tip, one point eight three meters out. I’m about to turn the propeller a quarter of a turn, so before I do, which of these two dots do you think turns through the bigger angle?

As the blade turns, both dots sweep out exactly the same ninety degrees, but the paths they leave behind are very different. Peel those two curved paths off and lay them flat on the lakebed, and the inner one is about ninety-six centimeters long while the tip’s is two point eight seven meters, three times longer, because the tip is three times farther out.

Three times farther out means three times the distance, and that one pattern sits behind a whole family of AP questions, so to write it down we need the way of measuring angles that physics likes best, radians. One radian is the angle whose arc along the rim is exactly as long as the radius. It’s as if you lifted the radius off and bent it around the rim, like a piece of string.

It takes a little over six of those, two pi of them, to get all the way around, on a circle of any size, so an angle in radians is just arc length divided by radius. Turn that around and you get delta s equals r delta theta, which says a point’s distance along its circle is its distance from the axis times the angle it turns through.

A quarter turn is pi over two radians, so the inner dot travels zero point six one meters times pi over two, about ninety-six centimeters, and the tip travels one point eight three meters times pi over two, about two point eight seven meters, just like the tapes. And since a radian is a length divided by a length, it quietly drops out, leaving plain meters.

Plug in ninety degrees instead and the tip travels a hundred sixty-five meters in a quarter turn, longer than a football field, which is your warning sign that the angle has to be in radians. In one full turn, two pi radians, the tip travels about eleven and a half meters, and this propeller turned about thirty-five times every second. All of this works because the propeller is a rigid system, so every point on it turns through the same angle in the same time, even though different points travel very different distances.

That angle has a sign, too. From in front of the airplane, the propeller in our model turns counterclockwise, but walk around behind it and the very same motion looks clockwise, so always say where you’re looking from and which way you call positive.

One spin, many speeds

Now let’s turn distances into speeds by dividing delta s equals r delta theta by the time it took, delta t, which turns the left side into the dot’s speed along its circle, v. On the right, r stays put, and delta theta over delta t is the angle turned per second, the angular velocity omega, so v equals r omega.

You might think the outer dot turns faster, because it travels so much farther. But every point on the blade shares one omega, so the outer dot spins at the same rate as the inner one and only moves faster because it has farther to go.

Here’s the real propeller at full speed with every point’s velocity drawn at one instant, each arrow at right angles to the blade and longer the farther out it sits. Look at the tips of those arrows, because they line up on one straight line through the center, and that straight line is a graph. Plot speed against distance from the axis and v equals r omega is a straight line through the origin with slope omega, while omega against r is perfectly flat, because every point shares it.

This exact picture, two points on one spinning object, is a classic AP question, and the trap it’s fishing for is the idea that the outer point spins faster. Point A is near the hub and point B is at the tip of the same propeller blade, so which statement correctly compares their angular speeds and their linear speeds?

Mach 1.18 on the ground

Let’s do the calculation an engineer would do, starting from the spec sheet, which says the propeller turned at two thousand one hundred three revolutions per minute and the tips sat about one point eight three meters from the center.

We want the tip speed, so our knowns are r and that rotation rate, and the equation that connects them is v equals r omega, once omega is in radians per second. Every revolution is two pi radians and every minute is sixty seconds, so two thousand one hundred three times two pi, divided by sixty, is about two hundred twenty radians per second.

Multiply by the radius, one point eight three meters times two hundred twenty radians per second, and you get about four hundred three meters per second at the tip, roughly nine hundred miles an hour. Sound travels at about three hundred forty meters per second at fifteen degrees Celsius, so the tip was moving about one point one eight times the speed of sound, which matches the published Mach number.

It passes the sanity checks, because the center, where r is zero, doesn’t move at all, and a point halfway out moves half as fast. And here’s a check you can do almost in your head. The tip covers about eleven and a half meters every turn, and the propeller makes about thirty-five turns every second.

So that’s about four hundred meters every second, the same answer by a completely different road, and that’s really what the two pi in the conversion was doing. Skip the conversion and multiply one point eight three by two thousand one hundred three, and you get almost thirty-nine hundred meters per second, more than eleven times the speed of sound, which is your cue to check the units.

The turbine behind the propeller turned much faster than the propeller did. It spun at fourteen thousand three hundred revolutions per minute, six point eight times as fast, so a gearbox sat between them and slowed the propeller down. If we bolted the propeller straight onto the turbine, omega would go up six point eight times, and so would every speed on the blade, putting the tips near eight times the speed of sound.

Now you try one. At the same rotation rate, how fast is a point one point two meters from the axis moving, in meters per second?

Where the sound barrier breaks

So the tip is supersonic and the hub isn’t, which means some point along the blade moves at exactly the speed of sound, and I’d love to know where. Call that distance r star. There, r omega equals c, the speed of sound, and dividing both sides by omega gives r star equals c over omega, before we’ve used a single number.

That already tells a story, because a bigger omega means a smaller r star, so spinning faster pulls the sound barrier in toward the hub and more of the blade goes supersonic. With numbers, three hundred forty divided by two hundred twenty is about one point five five meters, so at this rotation rate, in fifteen-degree air, only the outer twenty-eight centimeters of each blade, about eleven inches, moved faster than sound.

On a graph of speed against distance from the axis, the blade is the straight line through the origin. And the speed of sound is a flat line at three hundred forty meters per second. They cross right at r star, and every point to the right of that crossing is supersonic.

And here’s that answer on the real propeller, a ring one point five five meters from the center. Outside it the blade itself outruns sound, and in our model that’s where the shock waves come off. And that dimmer pink ring is the guess you made at the very start, so you can see how close you came, now that one line of algebra has told us exactly where the ring belongs.

Now I’m handing you the throttle, but work out your number first. What rotation rate puts the very tip right at the speed of sound? Once you have it, set the throttle there and watch whether the ring lands on the tip. Now put it back to full speed and change the weather, to a hot desert afternoon at thirty-five degrees Celsius, where sound travels at about three hundred fifty-two meters per second.

At the same rotation rate on that hot afternoon, what happens to the part of each blade that’s supersonic? The ring slides out to about one point six zero meters, so on a hot day only the last twenty-three centimeters break the sound barrier.

Spinning up

So far the propeller has turned at a steady rate, so let’s start it from rest and suppose it reaches full speed in twenty seconds. That twenty is our own number, for illustration, because we couldn’t find the real one anywhere. While it spins up, these new arrows show each dot’s tangential acceleration, pointing along its path and three times longer at the tip.

It’s the same story again. Start from v equals r omega, and any change in omega makes a change in v that’s r times as big, so dividing by the time gives a sub T equals r alpha.

Zero to two hundred twenty radians per second in twenty seconds is an alpha of eleven radians per second squared, the same for every point of the blade. So the tip gets about twenty meters per second squared, about two g, and the inner dot, a third as far out, gets a third as much, about six point seven.

At full speed alpha is zero, so the tangential acceleration is zero too, and yet the tip is still accelerating, because its velocity keeps changing direction as it goes around. That’s the centripetal acceleration from circular motion, pointing toward the axis, and at this tip it’s about nine thousand times the acceleration of gravity.

So every point on a propeller spinning at a perfectly steady rate is still accelerating, except the very center, and the only piece that drops to zero is the tangential one. When the engine shuts down, alpha reverses, so the tangential arrows swing around to point backward, against the motion, longest at the tip again. While the propeller winds down like this, how do the inner dot and the tip dot compare in angular acceleration and in tangential acceleration?

And vice versa

So far we’ve gone from the spin to the speed, and now let’s run the equation backwards, with two machines that look almost exactly alike and follow opposite rules. A record turns at exactly thirty-three and a third revolutions a minute, about three and a half radians per second, and the needle starts near the edge, about fourteen and a half centimeters out.

As the needle drifts inward, the rotation rate never changes. But the groove under the needle slows down, from about half a meter per second at the edge to about a fifth of a meter per second near the label. Unroll one turn of groove at each end and the outer one is ninety-two centimeters long while the inner one is only thirty-eight, yet both hold the same one point eight seconds of music.

Those squeezed-up wiggles are harder for the needle to trace, so bright sounds can pick up more distortion near the label, and that’s why mastering engineers like to put the brightest songs near the start of a side. Now stop the turntable. Every point on it slows at the same alpha, so the dot near the rim, one and a half times as far out, gets a backward arrow one and a half times as long.

A CD does the opposite. Its laser must see the track pass at one point two meters per second the whole time, and it reads from the inside out, so as the laser moves outward, the disc slows down.

That’s v equals r omega run backwards, omega equals v over r. The inner track sits twenty-five millimeters from the center, so omega is one point two divided by zero point zero two five, forty-eight radians per second, about four hundred sixty revolutions per minute. As the laser moves out to the edge, fifty-eight millimeters from the center, omega drops to about twenty-one radians per second, just under two hundred revolutions per minute, so a CD really does spin more slowly the longer it plays.

Now you drive the CD player, but do the math first. Work out how far from the center the laser has to sit for the disc to turn at exactly three hundred revolutions per minute, then slide it there and let go.

Now look at the CD at one single instant. Two dots on the disc share one omega, so the dot near the edge moves faster than the dot near the center, exactly like the propeller blade.

Now build the CD’s graph yourself. Slide the laser to three different spots and let go each time, and every point you drop on the graph is a different moment of the song.

And remember what those points mean, because each one is a different moment of the song, with the laser somewhere else, while at any one moment the whole disc still turns with a single omega. As graphs, the record keeps omega flat, so its speed is a straight line through the origin, like the propeller’s, while the CD keeps its speed flat, so its omega falls as the laser moves out, the very curve you just built.

Choose the tip speed first

Rotor designers usually run the equation backwards too, because they pick a tip speed they can live with first and work back to omega. Try it on Ingenuity, NASA’s little helicopter on Mars. Its blades reached sixty centimeters from the axis, with tips moving at about one hundred seventy meters per second. What was its angular speed in radians per second?

It couldn’t go much faster, because on Mars sound travels at only about two hundred forty meters per second, so Ingenuity’s tips were already at about seven tenths of the speed of sound there. Seventy years after the Thunderscreech, NASA went back to supersonic blade tips on purpose. In March of twenty twenty-six, it spun a new rotor for a future Mars helicopter in a test chamber at JPL until the blade tips broke the sound barrier, reaching Mach one point oh eight.

The line that earns the point

Let’s finish with a question the AP exam loves. Picture a string wound around a spool of radius two centimeters that turns on a fixed axle, and you pull fifty centimeters of string off it in one second, starting from rest and speeding up steadily. It asks how far the spool turns. The next sentence is the one that earns the point on the exam, so I’ll write it out in full.

Because the string doesn’t slip, the length of string that comes off equals the arc length traveled by a point on the rim, so l equals r delta theta, and that means delta theta equals l over r. So with that two-centimeter radius, how many radians does the spool turn while those fifty centimeters of string come off?

Now its steady speed-up gives us the rest. Starting from rest, the string averages half a meter per second, half its final speed, so it ends at one meter per second, and gaining that in one second means one meter per second squared. Divide each one by the two-centimeter radius, and the spool ends at an omega of fifty radians per second, with an alpha of fifty radians per second squared the whole way.

Now I want you to explain something, the way the exam asks you to. Spool B has twice the radius of spool A, and the same length of string comes off each, so which one turns through the bigger angle, and why? Read these three answers the way an AP grader would, and pick the one that earns the point.

That first answer, spool A is smaller, so it spins more, picks the right spool and still loses the point, because it never connects the length of string to the angle, and that connection is the physics. The answer that earns it says A turns through twice the angle, because the string doesn’t slip, so the length pulled off equals r delta theta, and with the same length and half the radius, A’s angle is twice B’s.

Everything agrees about the angle

So here’s what to carry away. Every point of a rigid spinning object turns through the same angle in the same time, sharing one omega and one alpha, and the farther out a point sits, the farther it travels and the faster it moves.

And while the spin rate is changing, each point’s tangential acceleration grows with r in the same way, while a steady spin makes it zero everywhere. That’s delta s equals r delta theta, v equals r omega and a sub T equals r alpha, with the angle in radians, and each one runs backwards just as well, omega equals v over r and delta theta equals l over r.

And now you know what was happening on that lakebed in nineteen fifty-five, where one omega put the sound barrier one point five five meters from the center, so on that parked airplane only about the last eleven inches of each blade went faster than sound.