Speed and Tension in a Vertical Circle

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. You whirl a stone on a string in a vertical circle. To reach the top, it must rise through two lengths of the string, right? So you give it just enough speed at the bottom to climb that height. Will it go all the way round? Most students say yes. Today we will see why it does not. This comes up often in JEE, so let's begin.

0:27Bob on a string of length l: speed from energy

Look at the figure. The bob goes round the centre O on a string of length l. At the lowest point, A, its speed is u; B is where the string is horizontal, and C is the top. The string has now turned through theta, this Greek letter, from the downward vertical. How high has the bob risen? The lowest point is l below the centre, and the bob is now l cos theta below it. So it has risen h equals l times one minus cos theta. Got it? Now, energy. This sign, like an upside down T, means perpendicular: the tension T is perpendicular to the velocity v, so it does no work. Only gravity does work, with m the mass and g the acceleration due to gravity. So half m v squared equals half m u squared, minus m g h. Clear? Cancel m, double everything, and put in h. v squared equals u squared, minus two g l times one minus cos theta: the speed at any point. Okay?

1:29Tension: the radial equation

Now, the tension. In the figure, the weight m g points down, and the tension T pulls towards the centre. Which part of m g acts along the string? Yes, exactly! m g cos theta, pointing away from the centre. So the net pull towards O is T minus m g cos theta, which must equal the centripetal force, m v squared over l. With me? So, T equals m v squared over l, plus m g cos theta: the tension anywhere, once you know the speed. Right? Now put in v squared from energy, and take m over l outside the brackets. Two g l cos theta, plus one more from m g cos theta, makes three. So T equals m over l, times u squared minus two g l plus three g l cos theta. Okay?

2:17Bottom, side and top

Now, three special points. At the bottom, theta is zero degrees, the small circle being the degree sign, and cos zero is one. So T equals m u squared over l, plus m g, the greatest tension in the circle. Clear? With the string horizontal, theta is ninety degrees, cos is zero, and T equals m u squared over l, minus two m g. Now you try the top: theta is one hundred and eighty degrees, cos is minus one. What multiple of m g do we subtract? Five m g! Very good! Minus two g l, minus three g l. So at the top, T equals m u squared over l, minus five m g, the least tension. Got it? Now a topper's trick. T bottom minus T top, where the small words below each T say where. The m u squared over l cancels, leaving six m g, whatever the speed, provided the bob goes all the way round. Remember it!

3:15Completing the circle

Now, back to our puzzle, at C, the top. The string can pull but never push, so the tension there must be zero or more. T equals m v top squared over l, minus m g, with v top the speed at the top, and this sign, greater than or equal to, says it is at least zero. So how small can v top be? Root g l! Correct! v top squared must be at least g l. The double arrow means so, and this is the root sign. Clear? Now, climbing two l needs half u squared equal to g times two l, so u equals root four g l. Is that enough to go round? No, no, careful! The bob rises two l, so v top squared equals u squared minus four g l. With u squared equal to four g l, that is zero, less than g l! So the string goes slack before the top, and the stone leaves the circle instead of going round, okay? So we need u squared minus four g l to be at least g l, which gives u at least root five g l. Root five, not root four: puzzle solved! Got it? At exactly root five g l, the critical speed, the top tension is zero. And at the bottom, five m g plus m g: six m g, just as our trick says. Right?

4:31Trap: a motor keeps the speed constant

Now, a real trap. Our six m g is for a bob going round freely under gravity. But what if a motor keeps the speed constant? Still six m g? No, careful! The radial equation still holds: T equals m v squared over l, plus m g cos theta, still least at the top. Okay? But now v is the same at top and bottom, so m v squared over l cancels. The m g cos theta term gives plus m g at the bottom and minus m g at the top, so the difference is two m g, not six m g. Clear?

5:06Example 1: a stone that just completes a vertical circle

Now, an easy one. There is a zero point two kilogram stone on a one metre string, just completing a vertical circle. Find its speeds and tensions at the top and bottom, with g as ten. Pause the video and try it. Done? Okay, let's check. Just completes means the critical case. At the top, v equals root g l, root ten. This wavy equals sign means approximately: about three point two metres per second, written m s to the power minus one. And the top tension is zero. Right? At the bottom, u equals root five g l, root fifty: about seven point one metres per second. Clear? And the bottom tension? Our trick: six m g. Six times zero point two times ten, the cross meaning times: twelve newtons, N for newtons. Quick, right?

6:00Example 2: tension at the bottom, side and top

Now a harder one. There is a zero point five kilogram bob on a two metre string, moving at ten metres per second at the lowest point. Find the tension at the bottom, with the string horizontal, and at the top. Again g is ten. Pause and try it. Done? Let's check. At the bottom, T equals m u squared over l, plus m g. Zero point five times one hundred, over two, is twenty five, plus five: thirty newtons. Clear? Now the string is horizontal. The bob has risen l, two metres, so v squared equals one hundred, minus two times ten times two: sixty. Now tell me, does the weight pull along the string here? No, it does not! Cos ninety degrees is zero, so the weight is at right angles to the string. T is just m v squared over l: zero point five times sixty, over two, fifteen newtons. Don't add m g here, okay? At the top, the rise is two l, four metres. So v squared equals one hundred, minus four times ten times two, which is twenty. And g l is also twenty! Interesting, right? So T equals zero point five times twenty over two, minus five: zero! The bob just completes the circle. And our trick checks it: thirty minus zero is thirty, which is six m g. Perfect!

7:22Quick revision: tension, root 5gl and the 6mg rule

So, quick revision. Speed from energy, tension from the radial equation. To go round, at least root g l at the top, so at least root five g l at the bottom. Going round freely, the bottom tension is six m g more than the top; at constant speed, only two m g more. Now try the practice problems on this page. All the best!