Kinematics of Rotation

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take a disc at rest, and let it speed up with a constant angular acceleration. Watch one point on its rim. At some moment, the total acceleration of that point makes forty five degrees with its velocity. How far has the disc turned by then? Surely that depends on the size of the disc, and on how fast it speeds up, right? Hold that thought. Today: the kinematics of rotation. It comes up often in JEE, and the rest of this unit builds on it. So, let's begin.

0:37Rigid body, fixed axis: one angle for all

So, first, a rigid body: the distance between any two particles never changes. No stretching, no bending, like a wheel or a door. Okay? Now let it rotate about a fixed axis, like a wheel on its axle. Every particle moves in a circle centred on the axis. Right? Now you tell me. One particle close to the axle, one on the rim. In one second, which turns through the bigger angle? The same angle! Yes, exactly. The body is rigid, so it turns as one piece: every particle turns through the same angle in the same time. So one set of angular quantities describes the whole body. Clear?

1:20The angular quantities

Now, the angular quantities. Theta, the Greek letter on screen, is the angular displacement, in radians, written rad. The angular velocity, omega, this curly letter, equals d theta over d t. Its unit is rad s to the power minus one, that is, radians per second. With me? If omega changes, we have the angular acceleration, alpha, another Greek letter. Alpha equals d omega over d t, in radians per second squared. And the second form, omega times d omega over d theta, has no time in it; it links omega directly with the angle. Okay? Angular velocity is a vector: see the small arrow on top of omega. It lies along the axis, by the right-hand rule: curl your right-hand fingers the way the body turns, and your thumb points along omega. Got it? Last, units. One revolution, written rev, is two pi radians. And n rpm, revolutions per minute, means omega equals two pi n over sixty, in radians per second. So never put rpm straight into a formula. Convert first. Right?

2:29Constant α: the same equations, new letters

Alright, now the most useful part. Suppose alpha stays constant. In a straight line, constant acceleration gave us v equals u plus a t, and its friends, remember? So what do we get for rotation? Exactly the same equations, just with new letters! First: omega equals omega nought plus alpha t. Omega nought, the omega with a small zero below, is the starting angular velocity. Okay? Second: theta equals omega nought t, plus half alpha t squared. Here theta is the angle turned from the start. With me? Third, with no time in it: omega squared equals omega nought squared, plus two alpha theta. Clear? And fourth, the angle turned in the n th second: theta, with n t h written below. It equals omega nought, plus alpha times, in brackets, n minus half. Right? Topper's trick: nothing new to learn. In the straight-line equations, replace s by theta, u by omega nought, v by omega, and a by alpha; that is what the arrows mean. But one condition: alpha must be constant. If alpha changes with time, go back to the definitions.

3:44A point at distance r from the axis

Got it? Now, a point on the body at a distance r from the axis. Its speed is v equals omega r: same omega, so the farther the point, the faster it moves. In vector form, v with its arrow equals omega cross r; the cross means the cross product, and r with an arrow is the point's position vector, measured from a point on the axis. Okay? If omega changes, the point speeds up or slows down along its circle: the tangential acceleration, a with a small t below, equals alpha r. Clear? Quick question. Say a wheel turns at a constant omega, so alpha is zero, and the tangential acceleration is zero. So a rim point has no acceleration at all, right? No, no, careful! It still moves in a circle, so its velocity keeps changing direction. That is the centripetal acceleration, a with a small c below, towards the axis. It equals omega squared r, which is also v squared over r. So zero alpha does not mean zero acceleration. With me? The two parts are at right angles, so the total acceleration is the square root of a t squared plus a c squared; that is the root sign on screen. Right?

4:58Example 1: angle turned in the 3rd and 5th seconds

Alright, Example one. There is a wheel with uniform angular acceleration. It turns through twenty radians in the third second, and twenty eight radians in the fifth second. Find omega nought and alpha. Pause the video and try it first. Done? Okay, let's check. Some students put t equals three into theta equals omega nought t plus half alpha t squared, and set it equal to twenty. Is that correct? No, no, wait! That is the angle in the first three seconds. The question says in the third second: only from two seconds to three seconds. So use the n th second formula with n equals three: omega nought plus alpha times, three minus half, equals twenty. The double arrow means so: omega nought plus two point five alpha equals twenty. Okay? Same for the fifth second, with n equals five: omega nought plus four point five alpha equals twenty eight. Right? Now subtract the first equation from the second. Omega nought cancels, and two alpha equals eight. So alpha equals four radians per second squared. Clear? Put it back: omega nought equals twenty, minus two point five times four; this cross just means times. Ten radians per second. Both answers, in four short lines. Perfect!

6:25Example 2: rim acceleration at 45 degrees to velocity

Now a harder one: our puzzle from the start! So, a disc starts from rest with constant angular acceleration. Through what angle has it turned when the total acceleration of a rim point makes forty five degrees with its velocity? The small circle means degrees. Pause the video and try it. Done? Okay. Many students say: alpha is not given, the radius is not given. So we are stuck, right? Not at all! Look at the directions. The velocity, v with its arrow, is along the tangent, and so is the tangential acceleration; it points the same way as the velocity, because the disc is speeding up. The centripetal acceleration points towards the centre, at right angles to the velocity. With me? Now, the total makes forty five degrees with the velocity exactly when the two parts are equal. So a c equals a t, that is, omega squared R equals alpha R, with capital R the radius of the disc. The radius cancels: omega squared equals alpha. Okay? Next, it started from rest, so omega nought is zero. Our third equation becomes omega squared equals two alpha theta. Right? Put the two together: two alpha theta equals alpha. Alpha cancels! So theta equals half a radian. That's it! See what happened? Alpha cancelled, and the radius cancelled. So the answer is independent of alpha and of the radius: every such disc, big or small, slow or fast, has turned through half a radian at that moment. Our puzzle, solved! Clear?

8:05Quick revision: rotation equations and v = omega r

So, quick revision. For a rigid body rotating about a fixed axis, one theta, one omega and one alpha describe every particle. With constant alpha, use the straight-line equations, with theta, omega nought, omega and alpha in place of s, u, v and a. And for a point at distance r: v equals omega r, a t equals alpha r, and a c equals omega squared r. Now go and try the practice problems on this page. All the best!