Loops, Humps and Dips on Tracks

Transcript

0:00Introduction

Hello everyone! Picture a roller coaster with a loop of radius R, okay? The car starts from rest on a smooth track. The top of the loop is two R high, so a start a little above two R should be enough to go round, right? Hmm, no! That car would fall off before it reaches the top! Today we see why, and we also look at humps and dips in the road. These come up often in JEE, so let's begin.

0:28Inside a loop: the track can only push

So, inside the loop. At the very top, the ball is upside down. Which way does the track push it there? Up, or down? Down, towards the centre! Yes, exactly. Inside a loop, the track always pushes inwards. Capital N is this push, the normal force. And the weight m g, with m the mass and g the acceleration due to gravity, also points down. With me? Together they supply the centripetal force, so N plus m g equals m v squared over R, with v the speed. The double arrow means so: N equals m v squared over R, minus m g. Okay? Now, if the ball is slower at the top, N gets smaller. So can N become negative? Can a track ever pull the ball? No, never! The track can only push. So N is greater than or equal to zero: that is the sign with a line under it. Clear? So m v squared over R is at least m g. Cancel m and take the square root: v top is at least root g R, with the square root sign over g R, okay? The word top below the v is a subscript: the speed at the top. Just like a string: a string can only pull, a track can only push.

1:45From rest: the least release height

Now, our roller coaster, in the figure. The ball starts from rest at the orange dot marked start, at height h above the dashed bottom line, and runs into the loop of radius R, right? At the top, the figure reminds us: v squared at least g R, as we just found. And look, the figure and the caption below it already give the answer: h min equals five over two R, where the small min below the h means minimum. So our job now is to see why. So why is a start at about two R not enough? Think about the speed at the top. From two R, all the energy goes into climbing, so the speed at the top would be about zero; the wavy equals sign means approximately. And zero is less than root g R, that is the less than sign. So the ball falls off before the top: that is the trap! Now let's find the right height. The track is smooth, so there is no friction, and mechanical energy is conserved. Right? At the start: potential energy m g h, and no kinetic energy. At the top, two R up: m g times two R, in brackets, plus half m v top squared. Set them equal. With me? Now use v top squared at least g R, and cancel m g. So h is at least two R plus half R, okay? So h min equals five over two R, two and a half times the radius, exactly as the figure says! The mass cancelled, so this holds for any mass, but only from rest on a smooth track. Puzzle solved!

3:18On a hump: the centre is below

Now flip the curve: here is a hump of radius R, and a car going over it. The car is outside the curve, so at the top the centre is below it. The road pushes up with N, and m g pulls down. With me? Towards the centre now means down. So m g minus N equals m v squared over R, which means N equals m g, minus m v squared over R, right? So N is less than m g: that is why you feel light going fast over a hump. Now the car goes faster and faster. What happens to N? It gets smaller! Correct! At v equals root g R, N becomes zero. Any faster, and the road would have to pull the car down. It cannot, so the car leaves the road! The sign after v is greater than. Got it?

4:05In a dip, and inside a tube

Now the opposite: a dip in the road. At the bottom of the dip, will the road push with more than m g, or less? More! Yes, exactly. The centre is now above the car, so N minus m g supplies m v squared over R, okay? So N equals m g plus m v squared over R, and this N is the apparent weight, and it is more than the true weight, m g. That is why you feel pressed into your seat in a dip. Right? One last case: a ball inside a tube. Its outer wall can push it inwards and its inner wall can push it outwards, so the tube can push either way, and even at the top it can hold the ball up. The ball only needs to reach the top: v top at least zero. So do not use root g R for a tube! Clear?

4:51Example 1: normal force on a car in a dip

Alright, let's try one. Say a car moves at twenty metres per second, m s to the power minus one on screen, through a dip of radius fifty metres. Find the normal force on it as a multiple of its weight, with g as ten. Pause the video and try it. Done? Okay, let's check. It is a dip, so we add: N equals m g plus m v squared over R, that is, m times the bracket: ten, plus four hundred over fifty. Four hundred over fifty is eight, so N equals eighteen m. With me? Now, the question wants a multiple of the weight. So is the answer eighteen times the weight? No, no, careful! The weight m g is ten m, since g is ten. So eighteen m is one point eight times ten m: N equals one point eight m g. Almost twice the car's weight. That's it!

5:49Example 2: ball in a loop released from 12 m

Now a harder one, back to our loop. Say a zero point five kilogram ball, k g on screen, is released from twelve metres on a smooth track, leading into a loop of radius four metres. Find its speed and the track's force on it at the top, with g as ten. Pause the video and try it. Done? Okay, let's check. First, five over two R: five over two, times four, is ten metres. The cross means times. Ten is less than twelve, so it goes round, with speed to spare. Now, the speed at the top. It fell from twelve metres, so v squared is two times ten times twelve, right? Oh no, careful! That is the speed at the bottom of the loop. The top is two R up, eight metres, so the ball drops only twelve minus eight: four metres. Okay? So v squared is two times ten times four: eighty. And v is root eighty, about eight point nine metres per second. Clear? Now the force. At the top, N equals m times the bracket: v squared over R, minus g. That is zero point five times, eighty over four, minus ten. With me? Eighty over four is twenty, minus ten is ten, times zero point five: five. So the track pushes with five newtons; the upright capital N here is the newton, okay? It is positive, so the ball stays on the track. That's it!

7:20Quick revision: loop, hump, dip and tube

So, quick revision. Inside a loop the track only pushes, so v top is at least root g R; from rest on a smooth track, h min is five over two R, clear? On a hump, N equals m g minus m v squared over R; in a dip, m g plus m v squared over R, right? In a tube, v top just needs to be at least zero. Now go and try the practice problems on this page. All the best!