When the String Slackens; Rods and Beads

Transcript

0:00Introduction

Hello everyone! Let's start with a puzzle. Take a stone on a string of length l. To reach the top of the circle, it must rise two l, and energy says a speed of two root g l at the bottom is just enough. So it goes all the way round, right? On a light rod, yes. But on a string, no! The string goes slack before the top, and the stone falls inside the circle. Why, and where exactly? This comes up often in JEE, so let's find out.

0:31Recap: tension at angle θ

So, a quick recap with the figure. The bob hangs from the centre O on a string of length l. Point A is the lowest point, with speed u; B is level with the centre; C is the highest point. Theta, the Greek letter, is the angle of the string from the downward vertical. The red arrow is the velocity, with speed v, at angle theta. On the right are last lesson's results: v squared equals u squared minus two g l times one minus cos theta; a complete circle needs u at least root five g l, where the tick shaped sign is the square root; and T at A minus T at C, with the small letters A and C below the T, equals six m g. On screen, the tension: T equals m v squared over l, plus m g cos theta. T is the arrow along the string, and m g, the weight, is the downward arrow. Okay? Now think. Can this string ever go slack while the bob is below the centre? No, it cannot! Below the centre, cos theta is greater than zero. So both terms are positive, and T is greater than zero: the string stays taut. With me? So the string can slacken only above the centre, where cos theta is negative. Clear?

1:44Three cases for a string

Now, three cases. To reach point B, level with the centre, the bob must rise a height l. So what speed u at the bottom just gets it there? Root two g l. Correct! So when u is less than or equal to root two g l, the bob never rises above the centre. It just oscillates, and the string stays taut. Got it? Now the middle case: u between root two g l and root five g l. Not enough to go round, so the bob just swings back like a pendulum, right? No, no, careful! This is the trap. The bob rises above the centre, and up there the tension drops to zero. The string goes slack! The two less than signs mean u lies strictly between the two values. Okay? And when u is greater than or equal to root five g l, the bob goes all the way round. Right?

2:35Where the string slackens

So where exactly? Put the speed formula into the tension formula. T becomes m over l, times u squared, minus two g l, plus three g l cos theta. The string goes slack where this becomes zero. With me? So cos theta equals two g l minus u squared, over three g l. In the middle case, u squared is more than two g l, so this is negative, and theta is more than ninety degrees. The small circle means degrees. Clear? Its height above the lowest point is l minus l cos theta. Simplify, and the height h is u squared plus g l, over three g, measured from point A, okay? Now, a common mistake. When the string goes slack, has the bob stopped? Not at all! Its speed there: v squared equals u squared minus two g l, over three, positive in the middle case. With the string slack, only gravity acts, so the bob moves as a projectile. Got it?

3:31Rods, tubes and beads: push as well as pull

Now, rods and beads. Why does a string fail? It can only pull, never push. So at the top, the bob needs v at least root g l, which means u at least root five g l. Okay? But a light rigid rod can push as well as pull. So can a tube, and a smooth ring with a bead sliding on it. Right? So think. With a rod, what is the least speed the bob needs at the top? Zero! Yes, exactly! The rod can push, so the bob only has to reach the top, with v greater than or equal to zero. To rise two l, u squared must be at least two g times two l. So u is at least two root g l. And that solves our puzzle! On a string, two root g l falls in the middle case, so the string slackens before the top. So in the exam, check first: string, or rod? Clear?

4:22Example 1: where the string slackens for u = root 3gl

Alright, let's try one. Here, a bob has speed root three g l at the lowest point. Where does the string slacken, and with what speed? Pause the video and try it. Done? Okay, let's check. First, which of our three cases is this? The middle case! Three g l lies between two g l and five g l. So the string will slacken. Right? Cos theta is two g l minus three g l, over three g l: minus one third. So theta is about one hundred and nine point five degrees. The wavy equals sign means approximately. Just above the centre. Okay? The height: three g l plus g l, over three g. That is four thirds of l above point A, which means l over three above the centre. Clear? And v squared is three g l minus two g l, over three: g l over three. So v is the root of g l over three. Not zero! Got it?

5:21Example 2: slack 30 degrees above horizontal, max height

Now a harder one, in two parts. Part one: what speed at the bottom makes the string slacken when it is thirty degrees above the horizontal? Part two: find the speed at that moment, and the greatest height above the centre that the bob reaches afterwards. Pause and try both parts. Done? Okay, let's check. The string is thirty degrees above the horizontal. So theta is thirty degrees, right? No, no, careful! Theta is measured from the lowest point: ninety degrees up to the horizontal, plus thirty, makes one hundred and twenty. Cos one twenty is minus one half, so minus one half equals two g l minus u squared, over three g l. That gives u equals the root of three point five g l. That's part one! Part two. v squared is three point five g l minus two g l, over three: zero point five g l. Clear? Where is the bob? Zero point five l above the centre, because l sine thirty is half of l. Its velocity is along the tangent, at right angles to the string, so sixty degrees above the horizontal. Okay? The string is slack now, so the bob is a projectile. Will it reach the top of the circle? Only the upward part of the velocity, v sine sixty, lifts it higher. So the further rise is v squared sine squared sixty, over two g. That is zero point five g l times zero point seven five, over two g: zero point one eight seven five l. Right? So the highest point is about zero point six nine l above the centre. But the top is l above the centre. So the bob never reaches the top! That's it!

7:02Quick revision: three cases, slack point, rods and beads

So, quick revision. On a string: up to root two g l, it swings; between root two g l and root five g l, it slackens above the centre; from root five g l, it goes round. The slack point is where T is zero, with cos theta equal to two g l minus u squared, over three g l; after that, a projectile. And a rod, a tube or a bead only has to reach the top: u at least two root g l. Now go and try the practice problems on this page. All the best!