Kinematics of Rolling

Transcript

0:00Introduction

Hello everyone! Picture a bicycle going past you. Here is a strange fact: the top of its wheel moves twice as fast as the bicycle. And the bit of tyre touching the road? At that instant, it is not moving at all! Same wheel, same moment. How? Strange, right? Today: the kinematics of rolling. By the end, you will see exactly why. Rolling comes up often in JEE, so let's begin.

0:30Rolling = translation + rotation

So, a rolling wheel does two things at once: it moves forward, and it spins. On the left of the figure: translation. Every point moves forward with the same velocity, v c m. That is v with the small subscript c m, the velocity of the centre of mass, labelled just v in the figure, okay? In the middle: rotation about the centre, with angular speed omega, the Greek letter on the curved arrow. Every rim point moves with speed omega times R, where R is the radius: forward at the top, backward at the bottom. Right? Rolling is these two added together. So tell me: what is the velocity of the point touching the ground? Translation gives it v c m forward; rotation gives it omega R backward. So the contact point moves forward with v c m minus omega R, clear? Now, rolling without slipping on a fixed surface means the contact point does not slide: its velocity is zero. So v c m equals omega R, the rolling condition. And since this holds at every instant, a c m, the acceleration of the centre, equals alpha R; alpha, another Greek letter, is the angular acceleration. Got it?

1:45When v is not equal to omega R, the body slips

Now, what if v c m is not equal to omega R? That is the equals sign with a slash through it, in the heading. Then the body slips. Suppose v c m is greater than omega R; which way does the contact point slide? Forward! Yes, exactly. v c m minus omega R is positive, so the bottom point is dragged forward. And when v c m is less than omega R, it slides backward. Keep this sign rule in mind for example two. With me?

2:15Top, centre and bottom

Now the right of the figure: real rolling. From here on, I write just v for v c m, so omega R is also v, okay? Now tell me, how fast is the top point moving? Two v! Correct. At the top, v forward plus omega R forward: two v, the top arrow. That is the first half of our puzzle: the top of the wheel really moves at twice the bicycle's speed. Right? The centre just moves with v. And the contact point, the orange dot P: v forward, omega R backward, so zero. The second half of the puzzle: the tyre on the road is at rest at that instant. Clear? Now a rim point at the height of the centre. Translation gives v forward; rotation gives omega R, which is v, straight up or straight down. So, what is its speed? Root two v! Yes, exactly. Two perpendicular parts, each v, so the speed is the square root of v squared plus v squared: root two v. Equal parts, so the direction is at forty five degrees to the horizontal; the small circle means degrees. In the figure, the back point moves forward and up; on the front side, forward and down. With me?

3:33Topper's trick: the instantaneous axis

Alright, now a topper's trick that shows all this at once. The contact point P is at rest, so at this instant the whole body is simply turning about P, okay? That is why the axis through P is called the instantaneous axis. So every point moves as if going round P, and its speed is omega times its distance from P, the contact point. And it moves perpendicular to the line joining it to P; that is the upside down T sign. Right? Let's test it. The dot means times. The top is two R from P, the full diameter, so its speed is omega times two R: two v. The level point is root two R from P, so its speed is root two v. Both match! In general, take a rim point at angle theta from P, measured at the centre; theta is the Greek letter. Its distance from P is a chord, two R sine theta over two. So its speed is omega times that: two v sine theta over two. Clear?

4:36Uniform rolling: accelerations

Now one last idea, and a classic trap. In uniform rolling, speed and spin stay constant, so a c m and alpha are both zero. Then every point has only a centripetal acceleration, v squared over R, towards the centre. Okay? So, the contact point. Its velocity is zero, so its acceleration is zero too, right? No, no, careful! Zero velocity at one instant does not mean zero acceleration. The contact point has acceleration v squared over R, upwards, towards the centre. Remember this one!

5:12Example 1: speeds of the top point and a level rim point

Now, example one, an easy one. Here, a wheel of radius zero point four metre rolls without slipping at eight metres per second; the minus one on the s means per second. Find omega, and the speeds of the top point and of a rim point level with the centre. Pause the video, and try it first. Done? Okay, let's check. No slipping, so omega equals v over R: eight over zero point four, twenty radians per second. Clear? Now the top. The wheel moves at eight metres per second, so the top does too, right? Oh no, careful! Remember the bicycle. The top moves at two v: two times eight, sixteen metres per second. Okay? Now the level point. Its parts are v, eight, forward, and omega R, twenty times zero point four, also eight, vertical. So the speed is the root of eight squared plus eight squared: eight root two, about eleven point three metres per second. The wavy equals sign means approximately. Right? On the front side, it points forward and down, at forty five degrees below the horizontal. Done!

6:25Example 2: is it rolling, and which way does friction act

Now, example two, a little harder. Here, a disc of radius zero point five metre moves on the ground with v c m four metres per second, and omega ten radians per second, spinning in the forward rolling sense. Is it rolling? Find the velocity of the contact point and the direction of kinetic friction. Pause the video, and try it. Done? Let's check. Test the rolling condition. Omega R is ten times zero point five: five metres per second. But v c m is four. Not equal! So it is not rolling; it is slipping. Right? The contact point moves with v c m minus omega R: four minus five, minus one metre per second. Minus means backwards: v c m is less than omega R, so it slides backwards, just as we said. Clear? Now friction. The disc moves forward, and friction opposes motion, so friction acts backwards, right? No, no, wait! This is a common JEE trap. Kinetic friction opposes the sliding of the contact point, not the motion of the disc. The contact point slides backwards, so friction acts forwards! Got it? And this forward friction speeds up the centre and slows down the spin, until v c m equals omega R, with me? That's it!

7:49Quick revision: v = omega R, top 2v, bottom at rest

So, quick revision. No slipping: the contact point is at rest, v c m equals omega R, and a c m equals alpha R, okay? Top two v, centre v, bottom zero, level point root two v. And the trick: the body turns about P, so speed is omega times the distance from P, right? Now go and try the practice problems on this page. You are ready for them. All the best!