Level Curves, Banked Roads and Railway Tracks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a road of smooth ice, with no friction at all. On a flat icy road, nothing pushes a car sideways, so it just slides straight on. So, can a car ever go round a curve with zero friction? Hold that thought. Today: level curves, banked roads and railway tracks. Questions on these come up often in JEE, so let's begin.

0:29Turning on a level road

Alright, first a level road, curving along a circle of radius R, okay? To go round at speed v, the car needs a force towards the centre: m v squared over R, with m its mass; the little two up on v means squared. On a level road, friction is the only horizontal force, so friction supplies it. With me? But friction has a limit: mu times N, at most. Mu, this Greek letter, is the coefficient of friction, and capital N, the normal reaction, equals m g here, with g the acceleration due to gravity. Right? So m v squared over R is less than or equal to mu m g; that is this sign, less than with a line under it. Then this double arrow means 'so': v max, the greatest speed, equals the square root of mu g R; the small max below the v labels it. Clear? Now tell me. On the same curve, must a heavy truck go slower than a small car? Many students say yes. But look! The mass m cancels, so v max does not depend on the mass at all, for the same mu. Got it?

1:39Banked road at the design speed

Now, the banked road. Look at the figure. The road is tilted at an angle theta, this Greek letter; the dashed line is the horizontal. The weight m g acts down, and the normal reaction N is perpendicular to the road. So, which way does N point now? Yes, exactly! It tilts inwards, towards the centre: see the arrow at the top. Say there is no friction, like on our icy road, so ignore the orange arrow f for now. Vertically, the upward part of N, N cos theta, cos for cosine, balances m g. Okay? And the horizontal part, N sin theta, sin for sine, points towards the centre. So it supplies m v squared over R, the force the car needs. The road itself turns the car! Divide the second line by the first: sine over cosine is tan, short for tangent, and N and m cancel. So tan theta equals v squared over R g, like the label on the figure. This v is called the design speed: no friction is needed. So yes, at the design speed, a car turns even on ice! With me? Now, on an incline, N equals m g cos theta. So here too, N equals m g cos theta, right? No, no, careful! That is for a block sliding along an incline. This car moves on a horizontal circle, so the vertical balance gives N equals m g over cos theta, more than the weight. Clear?

3:12Banked road with friction

Now, real roads have friction, with coefficient mu. If the car goes faster than the design speed, which way does it tend to slip: up the slope, or down? Up the slope, outwards. Correct! Friction opposes that, so it acts down the slope, like the orange arrow f. Its horizontal part points inwards, so it helps the turn. Okay? Below the design speed, it is the opposite: the car tends to slide down, so friction acts up the slope. Oh no, careful! Friction is not always up the slope; it depends on the speed. Got it? So there is a greatest speed. v max equals the square root of g R, times mu plus tan theta, over one minus mu tan theta. The plus on top is friction helping the turn. Right? And a least speed, v min, with small min below it: the square root of g R, times tan theta minus mu, over one plus mu tan theta. Topper's trick: learn one, and swap the signs for the other. Between these two speeds, the car does not slip. Clear?

4:18Railway tracks and the leaning cyclist

Now, trains. To bank a railway track, the outer rail is raised by a small height h. The distance between the rails, small l, is called the gauge. Okay? The angle is tiny, so h is about l tan theta; the wavy equals sign means approximately. And at the design speed, tan theta is v squared over R g, so h is about l v squared over R g. With me? Now, a cyclist on a level curve leans inwards. At what angle? Hint: think of banking. Yes, the same formula! He leans at an angle theta from the vertical, where tan theta equals v squared over R g. Right? But the road is level, so friction must supply m v squared over R, and friction is at most mu m g. So mu must be greater than or equal to tan theta; that is this sign, greater than with a line under it. Got it?

5:14Example 1: banking angle for 20 m/s with no friction

Now, Example one. The road has radius eighty metres, where m means metres, and is to be banked for twenty metres per second, written m s to the minus one, with no help from friction. Find the angle. Pause the video and try it. Done? Okay, let's check. No help from friction means twenty metres per second is the design speed, so tan theta equals v squared over R g. Take g as ten metres per second squared, written m s to the minus two. Okay? Now, twenty squared, over eighty times ten; the cross means times. That is four hundred over eight hundred, one half. Careful: forgetting to square the twenty is a common slip. Clear? So theta is tan inverse of one half: the minus one up on tan means inverse. That is about twenty six point six degrees; the small circle means degrees. And notice, we never needed the mass!

6:15Example 2: greatest and least speed on a 45 degree curve

Now a harder one, Example two. The curve has radius forty metres, banked at forty five degrees, and mu is zero point five. Find the greatest and least speeds without slipping. Pause the video and try it. Hint: tan forty five degrees is one. Done? Okay, let's check. With g as ten again, g R is ten times forty, four hundred, and tan forty five degrees is one. With me? Now, at the greatest speed, which way does friction act? Yes, down the slope, like the orange arrow! So mu plus tan theta goes on top: v max squared is four hundred times zero point five plus one, over one minus zero point five; the dot means times. That is four hundred times one point five over zero point five, twelve hundred. Okay? So v max is twenty root three, about thirty four point six metres per second. Right? Now the least speed: friction acts up the slope, so the signs swap. v min squared is four hundred times one minus zero point five, over one plus zero point five, which is four hundred times zero point five over one point five, four hundred over three. Got it? So v min is twenty over root three, about eleven point five metres per second. If you got both, very good!

7:41Quick revision: level road, banked road, rail and friction

So, quick revision. Level road: friction turns the car, and v max is the square root of mu g R, whatever the mass. Banked road: no friction is needed at the design speed, where tan theta equals v squared over R g; for rails, h is about l tan theta. With friction: down the slope above the design speed, up the slope below it. Now go and try the practice problems on this page. All the best!