Centripetal Acceleration

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a stone going round and round in a circle, at a perfectly steady speed. So, tell me: is this stone accelerating? Most students say, no acceleration, obviously. Hmm, are you sure? Today we settle it. And look, in JEE, almost every question on circular motion, from a stone on a circle to the earth going round the sun, starts with this one idea. So, let's begin.

0:32Constant speed, changing velocity

So, first, a name. Motion on a circle at constant speed is called uniform circular motion. Our stone is doing exactly this. Clear so far? Now think carefully. The speed is constant, okay. But is the velocity also constant? Not at all! Velocity is a vector, so it has a size and a direction; the little arrow on top of v means a vector, okay? At every point, v vector is along the tangent, the line that just touches the circle there. And as the stone goes round, the tangent keeps turning, right? So the size stays the same, but the direction keeps changing. And acceleration is the rate of change of velocity, in size or in direction. Here the direction changes, so v vector changes, so there is an acceleration; the double arrow means so, okay? So our stone is accelerating, all the time! That is half of our puzzle. Now the other half: which way does this acceleration point?

1:33It points towards the centre

Yes, towards the centre! Look at the left picture. The point O is the centre of the circle. The blue arrow, v vector, is the velocity, along the tangent. And the green arrow points from the particle straight towards O, along the dashed line, okay? That is the acceleration. The right picture, where the speed also changes, is for the next video. Acceleration towards the centre has a special name: centripetal acceleration, written a sub c; the small c below the a is for centripetal. And its size is a sub c equals v squared by r: the speed squared, divided by the radius r. The same formula is written next to the green arrow. Clear? So, in uniform circular motion, the total acceleration a is just a sub c, pointing towards the centre, as written under the left picture. Nothing along the tangent, because the speed does not change.

2:28One acceleration, three forms

Got it? Now, remember from the last video: v equals omega r. Omega, the Greek letter that looks like a curly w, is the angular speed. Put v equals omega r into v squared by r, and you get omega squared r. Or split it as omega into v. So, a sub c equals v squared by r, equals omega squared r, equals omega v. Pick the one that fits your data, okay? And the vector form: the acceleration vector, a vector, equals minus omega squared, times r vector. Here r vector is the position vector, from the centre to the particle. It points outwards, right? The minus sign flips it, so the acceleration vector points inwards, towards the centre. With me? Alright, quick question. Take a bigger circle, a bigger r. Does a sub c become bigger, or smaller? In v squared by r, the r is at the bottom, so smaller, right? Careful, this is a JEE trap! It depends on what stays the same. If the speed v is the same, then yes: a sub c is proportional to one by r; this sign means proportional to. Bigger circle, smaller acceleration. But if omega is the same, like for two points on one rotating disc, use omega squared r. Now a sub c is proportional to r. Bigger circle, bigger acceleration! So, first check: is v fixed, or is omega fixed? Then choose the form. Clear?

4:02How much does the velocity change?

Now, one more classic trap. Delta v vector means the change in velocity; the little triangle, delta, means change in. After half a revolution, the speed is the same as at the start. So delta v vector is zero, right? No, no, careful! After half a revolution, the particle is on the opposite side, moving the opposite way. So v vector becomes minus v vector; the arrow here means becomes, okay? So the change is final minus initial: minus v vector, minus v vector. Its size, mod delta v vector, is two v; the two straight bars mean the size. Not zero: two v! Got it? And after a quarter revolution, the tangent turns through a right angle. So v vector turns through ninety degrees; the small circle means degrees. Now we have two arrows, each of size v, at right angles. So the change has size root of, v squared plus v squared; the tick-shaped sign is the square root. That is root two v. Remember both: two v for half a turn, root two v for a quarter turn. We will use this in Example two.

5:13Example 1: acceleration of a stone on a circle

So, remember our stone from the start? Now it gets numbers. Here, a stone moves at two m s to the power minus one, that is, two metres per second, on a circle of radius zero point five m, half a metre. Find its acceleration. Pause the video, and try it first. Done? Okay, let's check. First, which form of our formula: v squared by r, or omega squared r? Yes, v squared by r! We know v and r directly. Two squared is four, and r is zero point five. So, four by zero point five. Four divided by half is eight. So the acceleration is eight metres per second squared, that is m s to the power minus two, directed towards the centre. Remember, the direction is part of the answer, so always write it. And our puzzle is solved: a steady speed, and still an acceleration of eight metres per second squared!

6:14Example 2: average vs centripetal acceleration, quarter turn

Now a harder one, JEE level. Here, a particle moves at ten metres per second on a circle of radius five metres. Find its average acceleration over a quarter revolution, and compare it with its centripetal acceleration. Pause and try. Hint: a quarter turn gives root two v. Done? Okay, let's check. First, the centripetal acceleration. v squared by r: ten squared is a hundred, by five, that is twenty metres per second squared. Now, the acceleration has size twenty at every instant. So the average over the quarter turn is also twenty, right? Hmm, let's not guess; let's calculate. Average acceleration is the change in velocity, divided by the time taken. In a quarter revolution, mod delta v vector is root two v. With v equal to ten, that is ten root two metres per second. Now the time t. The quarter arc is one fourth of two pi r, the full circle. With r equal to five, that is five pi by two metres. Divide by the speed, ten, and t is pi by four seconds. So the average acceleration, a with the small a v g below, meaning average, is mod delta v vector by t: ten root two, divided by pi by four. That is forty root two by pi. And that is about eighteen metres per second squared; the wavy equals sign means approximately. So the average is about eighteen, a little less than twenty. Not equal! The size of the acceleration stays twenty, but its direction keeps turning, so the average comes out smaller. Got it? This is a very common trap.

8:04Quick revision: v²/r = ω²r = ωv and change in velocity

So, quick revision. Constant speed, but a turning velocity, so the acceleration points to the centre. a sub c equals v squared by r, equals omega squared r, equals omega v. And the change in velocity: two v for half a turn, root two v for a quarter turn. Now go and try the practice problems on this page. All the best!