Mixed Problems: Set 3 (Circular Motion)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a car going round a curve on a level road, safely, with no skidding. Now the driver gets nervous and presses the brakes. Slower means safer, so now the car is even safer, right? Hmm, hold that thought. We will settle it by the end of this video. This is Set three of our mixed problems: circular motion. Questions on it come up often in JEE Main, and each one should take you only one to two minutes. So, one idea, a quick toolkit, and two problems. Let's begin.

0:39One idea behind all of Set 3

So, here is what is inside Set three. Fifteen problems: strings, springs, roads, railway tracks, turntables, rotors, and pendulums. It looks like seven different topics. But every one of them uses one single idea. Okay? Now tell me. Picture a body going round a circle at a constant speed. Is there a net force on it, or not? Yes, there is! The speed is constant, but the direction keeps changing. So the net force is m v squared over r, where m is the mass, v the speed, and r the radius. And this Greek letter is omega, the angular speed, so it is also m omega squared r. It points towards the centre. And even if the speed changes, the inward part is still m v squared over r. Clear? But this is not a new kind of force. Real forces supply it: tension, friction, N, the normal reaction, a spring force, or gravity. So here is the trick for this whole set: in every problem, first find who supplies m v squared over r. Got it? Then write two equations. This big Greek letter is sigma, and it means the sum. For a horizontal circle: the sum of the radial forces, F radial, equals m v squared over r, and the sum of the vertical forces is zero. With me? Now, here comes the trap. You draw the free-body diagram, and then add one more arrow, m v squared over r, pointing to the centre. Good idea, right? No, no, careful! In the ground frame, m v squared over r is the result of the real forces, not one more force. Add that arrow, and you count the same thing twice. Okay?

2:26Toolkit: level and banked roads

Alright, now the toolkit, starting with roads. The picture on the right is for banked roads; we come to it in a moment. First, a level road, no tilt at all. Only friction is horizontal, so friction supplies m v squared over capital R, the radius of the curve. Friction is at most mu times N, and here N is m g. Mu is this Greek letter, the coefficient of friction. And this sign means less than or equal to. With me? Now think. Take a heavy truck and a small car, on the same curve of the same road. Which one can go faster without skidding? Neither! They are the same. Cancel m on both sides: v max equals the square root of mu g capital R, okay? The max written below v means the greatest speed. The mass cancels, so it does not matter at all. Right? Now the picture: a road banked at the angle theta, this Greek letter. The blue arrow is N, the green arrow is m g, the weight, and the dark arrow on top points towards the centre. At the design speed no friction is needed: the part of N towards the centre does the job. That gives tan theta equals v squared over capital R g, the same line as in the picture. Clear? And with friction: faster than the design speed, friction f, this orange arrow, acts down the slope. Slower than the design speed, it acts up the slope. Got it?

3:54Toolkit: turntables, and changing speed

Next, a coin on a turntable, like a record. Friction supplies m omega squared r, where r is the coin's distance from the axis. Friction is at most mu m g, so mu must be at least omega squared r over g. This sign means greater than or equal to. Okay? Now, two coins on the same turntable, with the same mu, one near the axis and one farther out. The turntable speeds up slowly. Which coin slips first? The outer one, yes! Bigger r needs more friction, and both coins have the same limit, mu g. So the outer coin runs out of grip first. Right? And one last tool. If the speed changes on a curve, friction also supplies m a t along the path. That is a, with a small t written below, the tangential acceleration. We need it very soon. With me?

4:49Example 1: greatest safe speed on a level curve

Alright, two problems from Set three, an easy one first. Read it with me: find the greatest safe speed on a level curve of radius eighty metres, with mu equal to zero point five. Pause the video, and try it. One to two minutes! Done? Okay, let's check. First, find who supplies the m v squared over capital R that we need. On a level road, only friction, and friction is at most mu m g. Right? So v max equals the square root of mu g capital R, okay? With g as ten metres per second squared, that is the root of zero point five times ten times eighty. Clear? Zero point five times ten is five, and five times eighty is four hundred. The square root of four hundred is twenty. Twenty metres per second; on screen, m s to the power minus one. And see, we never needed the mass of the car! Very good.

5:48Example 2: least mu for a car braking on a curve

Now the hard one, and it is our puzzle from the start. Read it with me: a car at ten metres per second, on a level curve of radius fifty metres, brakes at one point five metres per second squared. On screen, m s to the power minus two means metres per second squared. Find the least mu that prevents skidding at that instant. Pause the video, and try it. Done? Okay, let's check. First, the radial part, the turning part: v squared over capital R, towards the centre. Ten squared is a hundred, over fifty: two metres per second squared. Okay? So friction must give two metres per second squared, and mu is two over ten, zero point two. Done, right? No, no, careful! That is the trap. The car is also braking. So friction has a second job: slowing the car along the road, a t equals one point five metres per second squared. Got it? Now, one part points towards the centre, and the other along the road, at right angles to it. How do we combine them? Pythagoras, yes, exactly! The square root of one point five squared plus two squared: two point two five plus four is six point two five, and its square root is two point five metres per second squared. Clear? That is the friction needed per unit mass, and it can be at most mu g. So mu must be at least two point five over ten: zero point two five. Now our puzzle. Without the brakes, zero point two was enough. With the brakes, we need zero point two five. So at that instant, braking needs more grip, not less. On a road with mu between these two values, braking this hard makes the car skid!

7:34Quick revision: level road, banked road, changing speed

So, quick revision. Every circle: find who supplies m v squared over r, and never draw it as an extra force. Level road: v max is the root of mu g capital R; banked road with no friction: tan theta equals v squared over capital R g, okay? And when the speed changes, friction supplies both parts, added at right angles. Now try the rest of Set three with a timer running, one to two minutes each. Then try the practice problems on this page. You can do this. All the best!