Rotating Frames and Speeding Up on a Curve

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You are sitting in a car, and it takes a sharp left turn. You slide to the right, towards the door, right? But wait. Nobody touched you. So which force threw you outwards? Today we answer this in two ways, and both are correct. Then we see why a car that speeds up on a curve skids sooner than you expect. Questions like these come up often in JEE, so let's begin.

0:30Two frames, one turning car

So, first stand outside on the road and watch the car turn. This is the ground frame. From here, tell me: who pushed you towards the door? Nobody! Yes, exactly. The ground frame is an inertial frame, so we use real forces only. Your body simply tends to go straight on. That is inertia. Okay? Meanwhile, the car turns left under you. So you slide right, until the door or the seat pushes you towards the centre of the turn. That real push makes you turn with the car. With me? Now, sit back inside the car. The car frame turns, so it is a rotating frame, rotating at omega. Omega is this Greek letter, the angular speed. In this frame you feel an outward centrifugal pseudo force, m omega squared r, where m is your mass and r is your distance from the centre of the turn. Clear? And a body at rest in this frame? The real forces balance the pseudo force. The door pushes you in, the centrifugal force pushes you out, and you stay in your seat. Got it? Quick question. In the ground frame, should we also draw this outward m omega squared r? No, no, never! In the ground frame there is no centrifugal force, just real forces and the true acceleration. Pick one frame, and never use both in one equation. Right?

1:54Why the earth's spin reduces g

Same idea, bigger scale: the earth. The earth spins about its axis, so a body on the equator moves on a circle of radius capital R, the earth's radius. Okay? In the earth's frame, a rotating frame, this body feels a centrifugal pseudo force, m omega squared capital R, outwards, while gravity pulls it inwards. Opposite directions, right? So tell me: does the spin make the effective g bigger, or smaller? Smaller, correct! g effective equals g minus omega squared capital R, okay? Here g is the acceleration due to gravity, and the small e f f below it means effective. How big is this drop? Example one will tell us.

2:39Speeding up on a curve: friction has two jobs

Alright, the second big idea. On a level curve at constant speed, friction points straight to the centre. But now the car is speeding up. Does friction still point straight to the centre? No! Now friction has two jobs. When the speed changes on a circle, the net force has a tangential part, m a t, along the motion, which changes the speed; the small t means tangential. And a radial part, m v squared over r, which turns the car, with v the speed and r the radius. Okay? These two parts are at right angles. So by Pythagoras, the net force F is m times the square root of a t squared, plus v squared over r, whole squared. The long tick is the square root sign, and the big brackets mean we square the whole of v squared over r. With me? Now, on a level road, friction supplies both parts, up to a limit: mu times N, okay? Mu is this Greek letter, the coefficient of friction, and N is the normal reaction. On a level road, N equals m g, so friction can give at most mu m g. Right? So the car stays on its path while this net force is less than or equal to mu m g. This sign, less than with a line under it, means less than or equal to. Past that, the car skids. Clear? Now a topper's trick. Look, m is on both sides, so it cancels. Just check: root of a t squared plus v squared over r whole squared, less than or equal to mu g. No mass needed! Got it?

4:17Example 1: how much the earth's spin reduces g

Now, example one. Read it with me. By how much does the earth's spin reduce the effective g at the equator? Capital R, the earth's radius, is six point four times ten to the power six metres, and one rotation takes twenty four hours. Pause the video, and try it first. Done? Okay, let's check. We just saw: in the earth's frame, g falls by omega squared capital R, okay? So we only need omega. One rotation takes twenty four hours. So we just put in twenty four, right? Oh no, careful! Omega must be in radians per second, so the time must be in seconds: twenty four times three thousand six hundred, eighty six thousand four hundred seconds. One rotation is two pi radians; pi is this Greek letter. So omega is two pi over eighty six thousand four hundred, about seven point two seven times ten to the power minus five; this wavy equals sign means approximately. The unit, rad s to the power minus one, means radians per second. Got it? Now square omega, and multiply by capital R, okay? Seven point two seven times ten to the power minus five, whole squared, times six point four times ten to the power six. Right? That gives about zero point zero three four metres per second squared, written m s to the power minus two. So the spin reduces g only a little, but it does reduce it. Very good!

5:50Example 2: speed at which a speeding-up car skids

Now a harder one, JEE level. Read it with me: a car starts from rest on a level circular track of radius one hundred metres, and speeds up at two metres per second squared. Mu, the coefficient of friction, is zero point four. At what speed does it begin to skid? Pause the video, and try it. Done? Okay. Many students say: level road, so v squared over capital R, the track radius, equals mu g. Mu g is four, so v squared is four times one hundred, four hundred, and v is twenty metres per second. Right? No, no, careful! The car is also speeding up, so friction has two jobs: the tangential part, a t equals two, and the radial part, v squared over R, together. Root of a t squared plus v squared over R whole squared must be at most mu g. Taking g as ten, mu g is zero point four times ten, which is four. Okay? The car begins to skid when the two sides are equal. Square both sides: v squared over one hundred, whole squared, equals sixteen minus two squared, which is twelve. Right? Take the square root: v squared over one hundred is root twelve, that is two root three. The double arrow means implies: v squared equals two hundred root three. With me? So v is the square root of two hundred root three. Two hundred root three is about three hundred and forty six, so v is about eighteen point six metres per second. See, less than the twenty from the trap! Part of the friction is busy speeding the car up, so less is left for turning, and it skids sooner, just as I promised at the start. Very good!

7:35Quick revision: centrifugal force, g at equator, skidding

So, quick revision. In a rotating frame, add an outward centrifugal pseudo force, m omega squared r, but never in the ground frame. The earth's spin makes g at the equator fall by omega squared capital R, okay? And when a car speeds up on a level curve, there is no skid while root of a t squared plus v squared over r whole squared is at most mu g. Now go and try the practice problems on this page. All the best!