Constraint Relations

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Block A sits on a smooth table, and its string runs over a pulley at the edge to hold up a hanging block B, okay? Now block A slides forward. How far does block B go down? Just as far, of course. Same string, right? Well, not always! Put a movable pulley in between, and block B goes only half as far. Today we learn these links between motions, called constraint relations. Pulley and wedge problems come up often in JEE, and they start right here. So, let's begin.

0:39The rule: the string's length is fixed

So, where do these links come from? The string itself. Take a light string that does not stretch, and stays taut. Its length, capital L, is fixed, however the bodies move. Okay? Step one: write L using the positions of the bodies the string touches. Each straight piece is just a distance between two of them. Right? Now, L never changes with time. So what is its rate of change? Zero, exactly! So, d L over d t equals zero; d over d t means the rate of change with time, small t. This links the velocities, and differentiating once more links the accelerations. That is the whole method. Clear?

1:23The movable pulley: half as far

Now look at the figure. Block A is on a table. Its string goes over a pulley at the edge, down under a movable pulley holding block B, and up to the ceiling. The arrow marked x shows block A moving a distance x, and the arrow marked x over two shows block B moving down. The green arrows marked T are the tension in the two strands. With me? See, block A moves x, but block B only x over two. One single string! So why only half? Use our rule: the total length is fixed. Block A moves x towards the edge, so the flat part of the string gets shorter by x. Okay? That length goes into the two strands on the movable pulley: together, they get longer by x. Right? Both strands hang from the same pulley, so each gets longer by the same amount, x over two. So the pulley, and block B with it, goes down x over two. That's it! Answering x is the trap: one string, yes, but two strands on the movable pulley. Divide by the time, and you get half the speed too. The same for accelerations: a sub B equals a sub A over two, where a sub A and a sub B are the accelerations of the two blocks. Got it? One more thing. With a light string over light, frictionless pulleys, both strands carry the same tension T, and both pull the movable pulley up. So the pulley feels two T upwards. Remember this one.

2:53Topper's trick: the average rule

Now, a trick toppers use. Every movable pulley moves with the average of the two ends of its string. So, a sub pulley equals a sub one plus a sub two, over two, where a sub one and a sub two are the accelerations of those two ends. Okay? Test it on our figure. One end is tied to the ceiling, and the other end, at the fixed pulley, moves just as fast as block A, right? So what is the average? Yes, exactly! Zero plus a, over two, which is a over two; here small a is the acceleration of the free end. The half rule again, in one line. Nice, isn't it? And the rule works for velocities too. But the two ends may move in opposite directions, so choose one direction as positive, and keep the signs. Example one will show why. With me?

3:43Bodies in contact: the common normal

Strings are not the only constraint. Sometimes rigid bodies simply touch: a rod resting on a wedge, or a block on a block. Okay? At the point of contact, draw the line perpendicular to the touching surfaces. That is the common normal. Got it? Now, if one body moved faster than the other along this normal, what would happen? They would fly apart, or push into each other! So, while they stay in contact, their velocity components along the common normal are equal. The solved examples on this page include a rod on a wedge. Clear?

4:17Example 1: velocity of a movable pulley, with signs

Now, example one. The two ends of a string over a light pulley are moving: one end down at three metres per second, the other end up at one metre per second. On screen, m s to the power minus one means metres per second. Find the velocity of the pulley. Pause the video, and try it first. Done? Okay, let's check. The average of three and one is two, so the pulley moves at two metres per second, right? No, no, careful! The ends move in opposite directions. Take up as positive. So v sub one, the velocity of the end moving down, is minus three, and v sub two, the end moving up, is plus one. Okay? Now average them: v sub pulley, the velocity of the pulley, equals minus three plus one, over two. That is minus two over two: minus one metre per second. Clear? The minus sign means downwards. So the pulley moves at one metre per second, downwards. Not two! Signs first, then the average. Got it?

5:24Example 2: block on a table pulled via a movable pulley

Now a harder one, JEE level: our puzzle, with numbers. Same figure. Block A, two kilograms, is on a smooth table; k g on screen is kilogram. Its string runs over a pulley at the edge, down under a movable pulley carrying block B, four kilograms, and up to a hook on the ceiling. Find the accelerations and the tension. Pause and try. Hint: you need three equations. Done? Okay, let's check. Three unknowns: the tension T, a sub A and a sub B, okay? Newton's second law gives one equation per block, so we need one more: the constraint! We already have it: a sub B equals a sub A over two. Right? First, block A, on the smooth table: the only horizontal force on it is the tension. So, T equals two times a sub A, clear? Now block B, with its pulley, which we take as light. Its weight pulls it down. What pulls it up? Just T? No, two T! Both strands pull up, remember? That is the classic trap. So, four g minus two T equals four a sub B, with g, the acceleration due to gravity, as ten metres per second squared; that is the m s to the power minus two on screen. With me? Four g is forty, and four a sub B is two a sub A, since block B has half the acceleration. The double arrow means implies: put in T equals two a sub A, and two T becomes four a sub A, so forty minus four a sub A equals two a sub A, okay? So six a sub A is forty: a sub A is twenty over three metres per second squared, and a sub B, half of that, is ten over three. Right? And T equals two a sub A: forty over three newtons; capital N is the newton. Block B has exactly half the acceleration of block A, our puzzle with real numbers! Constraint first, then Newton's law for each body. Got it?

7:31Quick revision: fixed length, average rule, common normal

So, quick revision. One: the string's length is fixed, so d L over d t is zero, and that links the motions. Two: a movable pulley moves with the average of its two ends; with one end fixed, that is half, and it feels two T upwards. Three: for rigid bodies in contact, the velocity components along the common normal are equal. Now go and try the practice problems on this page. All the best!