Non-uniform Circular Motion

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a car going round a curve at constant speed. Its acceleration points towards the centre of the curve; we saw that with centripetal acceleration. Now the driver presses the brake, while still on the curve. Where does the acceleration point now? Still towards the centre? Or straight backwards? Hmm, think about it. By the end of this video, you will know its direction, and its size too. And remember, in JEE, whenever the speed changes on a circle, this is the idea they test. So, let's begin.

0:38Speed changing: two parts of the acceleration

So, look at the figure. On the left, uniform circular motion: the particle goes round the centre O at constant speed. Its velocity, v vector, is the blue arrow along the tangent; the little arrow on top means a vector, okay? And the green arrow, pointing to the centre, is a sub c, equal to v squared by r, where r is the radius. The small c below a stands for centripetal. Now, on the right circle, the speed is changing. Is the green arrow still the whole story? No, something more is needed! The speed itself is changing, so there must be an acceleration along the velocity. That is the tangential acceleration, a sub t, the small orange arrow along v vector. The small t below a stands for tangential. So, a sub t equals d v by d t, the rate of change of speed, and that equals r alpha: the radius into alpha, the angular acceleration. Clear? And the velocity is still turning, right? So the centripetal part stays: a sub c equals v squared by r, towards the centre. Beside the figure, you also see it written as omega squared r, from our last lesson. Just the same thing. Now we have two arrows: a sub t along the tangent, and a sub c along the radius. What is the angle between them? Yes, ninety degrees! The tangent is always perpendicular to the radius. So the total acceleration, a vector, the dark arrow, is the hypotenuse of a right triangle. Its size is the square root of a sub t squared plus a sub c squared: the root sign on screen, okay? Look, never just add a sub t and a sub c. They are perpendicular. That is the first JEE trap.

2:26Direction: angle between acceleration and velocity

Alright, we have the size. Now the direction. We measure it from the velocity. Let phi, the Greek letter on screen, be the angle between a vector and v vector. With me? Look at the right triangle again. Here, a sub t lies along v, next to the angle phi, and a sub c is the side opposite to it. So, tan phi equals what? Yes, a sub c by a sub t! The opposite side by the adjacent side. So, tan phi equals a sub c by a sub t. Got it? Now, when the particle speeds up, a sub t points along v, just like the orange arrow in the figure. So the total acceleration leans forward, and phi is less than ninety degrees. The little circle means degrees, okay? Now, back to our car. The driver brakes on the curve, so the speed falls. Which way does the acceleration lean now? Backwards, exactly! While slowing down, a sub t points opposite to v. So the total acceleration leans back, and phi is greater than ninety degrees. Not towards the centre, not straight back, but in between, leaning back. Our puzzle is half solved; the size comes in Example one.

3:41When the arc length s(t) is given

Now, one more situation. Sometimes JEE gives you the arc length s, the distance moved along the circle, as a function of time: s of t. How will you get the speed from it? Differentiate, of course! So, v equals s dot. The dot on top of s means d by d t, the rate of change with time. So, v equals d s by d t. Differentiate once more: a sub t equals s double dot, with two dots, which is d v by d t. But careful! This s double dot gives only the tangential part, not the whole acceleration. Students lose marks here. So, the full recipe. Find v, then a sub t, then a sub c equals v squared by r, with the speed at that instant, and finally combine them with the square root. Four steps, every time. Clear?

4:39Example 1: car braking on a curve, total acceleration

Alright, Example one, and it is our braking car! Here, a car moving at twenty m s to the power minus one, that is, twenty metres per second, on a curve of radius one hundred metres, brakes with a tangential retardation of three m s to the power minus two, that is, three metres per second squared. Find its acceleration. Pause the video, and try it first. Done? Okay, let's check. First, the centripetal part: a sub c equals v squared by r. Twenty squared, by one hundred: four metres per second squared. Right? Next, the tangential part. Retardation means slowing down, so a sub t is three metres per second squared, pointing backwards, against the velocity. For the size, this direction does not matter. Now the total. Three plus four: seven metres per second squared, right? No, no, careful! That is the trap. The two parts are perpendicular, remember? So, the total equals the root of three squared plus four squared: root twenty five, that is, five metres per second squared. And the direction: the car is braking, so the acceleration leans back, and phi is greater than ninety degrees. And there it is, puzzle solved! Five metres per second squared, leaning back from the velocity, not towards the centre.

6:07Example 2: speed and acceleration from s = t² + 2t

Now a harder one, JEE style. Here, a particle moves on a circle of radius three metres, and its arc length s equals t squared plus two t, in metres. Find its speed, its tangential acceleration, and its total acceleration at t equals two seconds. Pause and try. Hint: use the four-step recipe. Done? Okay, let's check. Step one: v equals s dot. Differentiate t squared plus two t, and you get two t plus two. At t equals two seconds, v equals six metres per second. Step two: differentiate again, and s double dot is two. So the acceleration is two metres per second squared. Done, right? Oh no, careful! Remember the warning. This s double dot is only a sub t, the tangential part. So, a sub t equals two metres per second squared. That answers the second part, but the total acceleration is still waiting. Step three, the centripetal part: v squared by r. Six squared is thirty six, and thirty six by three is twelve metres per second squared. Step four: combine. Root of two squared plus twelve squared: root of one hundred and forty eight. The wavy equals sign means approximately: about twelve point two metres per second squared. So that answer of two was way off. Always find both parts, then combine. Got it?

7:40Quick revision: total acceleration, never just add

So, quick revision. When the speed changes, there are two parts: a sub t equals d v by d t, which is r alpha, along the velocity, and a sub c equals v squared by r, towards the centre. Combine them with the square root, and tan phi equals a sub c by a sub t; never just add them. And if s of t is given, v is s dot, a sub t is s double dot, and the centripetal part still counts. And that braking car puzzle is now yours, size and direction both. You have got this! Now go and try the practice problems on this page. All the best!