Springs and Sudden Changes

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Block A hangs from a spring fixed to the ceiling, and block B hangs from block A by a string. Everything is at rest. Now I cut the string. Block B falls, of course. But what about block A, just after the cut? Does it stay at rest for that moment, does it fall too, or does it go up? Hold on to your answer. Today we learn what can change in an instant, and what cannot. In JEE, cut-the-string questions like this come up often. So, let's begin.

0:36What can change in an instant?

So, first recall the spring force from the last chapter. F equals k times x. Here k is the spring constant, the stiffness of the spring, and x is its extension, or its compression. So the force depends on how much the spring is stretched. Okay? Now think. To change x, the end of the spring must move, and that end is tied to a block at rest. So, can x change in zero time? No, it cannot! Moving takes time. So x cannot change in an instant, and the double arrow means "so": the spring force cannot jump. So just after a string is cut, every spring that is still attached pulls exactly as it did before. Clear? Now, what about a string? Can its tension jump in an instant? Yes, it can! The tension, capital T, and the normal force from a surface, capital N, are not tied to any stretch. They adjust themselves at once to whatever is needed, so they can jump. With me? And when a string is cut, its tension becomes zero at once. The string simply pulls no more. Right? So here is the method. Just after the cut, keep the force of every spring that is still attached as it was, remove what was cut, and find each acceleration, small a, from the forces that remain, one body at a time. Got it?

2:01Back to the puzzle

Now, back to our puzzle. Block A hangs from the spring, and block B hangs from A by a string, all at rest. Their masses are m A and m B; the small A and B written below the m are just labels. Okay? Before the cut, everything is at rest. So, how much force must the spring pull up with? Yes, exactly! It holds up both blocks. So the spring force is m A plus m B, in brackets, times g, the acceleration due to gravity. Right? Now the string is cut. Which forces act on block B just after? Only its weight, m B g! The string pulls no more. So, double arrow, a B equals g, downwards. Clear? Now the top block. Just after the cut, the velocity of A is still zero. So its acceleration is zero too, right? No, no, careful! Zero velocity does not mean zero acceleration. The extension has not changed, so the spring still pulls A up with m A plus m B, times g. But now only the weight of A, m A g, pulls down. So, m A plus m B, times g, minus m A g, equals m A times a A, where a A is its upward acceleration. With me? The m A g cancels, and a A equals m B over m A, times g, upwards. So, puzzle solved! Block A neither stays at rest nor falls; it jumps upwards. Got it?

3:34The puzzle in one picture

Here is the whole story in one picture. On the left, before: the spring force is m A plus m B, times g. On the right, just after the cut: the dashed line is the cut string, and the spring force is unchanged. The green arrow: B falls with g. The blue arrow: A goes up with m B over m A, times g. Okay? Now put in numbers from the course. Block A is two kilograms, block B is three kilograms. Take g as ten. How fast does A go up? Yes! m B over m A is three over two, and three over two, times ten, is fifteen metres per second squared, upwards! The m s to the minus two means metres per second squared. That is more than g, because block B is heavier than block A, right? And block B, of course, falls with g. That's it!

4:29Example 1: two springs hold a block, one spring is cut

Now, example one. Two vertical springs hang side by side, with spring constants one hundred and three hundred N m to the minus one, that is, newtons per metre. Then, a two kilogram block hangs at rest from both. The stiffer spring is cut. Find the acceleration just after; take g as ten m s to the minus two, that is, metres per second squared. Pause the video, and try it. Done? Okay, let's check. One spring is cut, so the block simply falls with g, right? No, no, wait! The other spring is still attached, still pulling. So first, before the cut. Side by side, both springs have the same extension x. Together they hold the weight: one hundred plus three hundred, times x, equals two times ten, twenty newtons. So x equals twenty over four hundred: zero point zero five metres. Okay? Each spring force is k times x. One hundred times zero point zero five is five newtons; three hundred times zero point zero five is fifteen newtons. The capital N after each number means newtons. Right? Now cut the stiffer spring. Which forces remain on the block just after? Its weight, twenty newtons down, and the softer spring, still five newtons up, because its extension has not changed. So the net force is twenty minus five: fifteen newtons, downwards. Clear? So a equals fifteen over two: seven point five metres per second squared, downwards. Less than g, because the soft spring still holds. That's it! Now look at those fifteen newtons. They are exactly the force the cut spring was giving, reversed. So here is a topper's trick. When a body at rest loses one force, and the other forces stay unchanged, the net force equals the lost force, reversed. Keep this for the next example. Okay?

6:32Example 2: ball held by a spring and a string, string cut

Now a harder one, JEE level. Here, a ball of mass m is held at rest by a horizontal string and a spring. The spring makes sixty degrees with the vertical; the small circle after sixty means degrees. The string is cut. Find the acceleration just after. Then, what if the spring were replaced by a string? Pause and try: the spring case first, then the string. Done? Okay, let's check. Before the cut, let capital S be the spring force. Its vertical part, S cos sixty degrees, balances the weight m g; its horizontal part, S sine sixty degrees, balances the tension, capital T, in the string. With me? Divide the two: T equals m g tan sixty degrees. Tan sixty is root three, with the square root sign you see, so T is root three m g. Okay? Now cut the string. The extension of the spring cannot change in an instant. So, which force disappears? Only T! The spring force and the weight stay as they were. So, our trick: the net force is T reversed, root three m g. Then a equals root three m g over m: root three g, horizontally, away from the side of the cut string. Got it? Now the second part: replace the spring by a string. Same answer, root three g, right? Oh no, careful! Now the other support is also a string, and its tension can jump. So the trick fails here. The ball starts to swing on a circle, from rest. Its velocity, v, is zero, so it has only tangential acceleration, along the circle. Clear? Along the circle, perpendicular to the string, only the weight has a part: m g sine sixty degrees. So a equals g sine sixty degrees: root three over two, times g, perpendicular to the string. Spring, root three g; string, root three over two g. Same setup, different answers. See the difference?

8:42Quick revision: spring force cannot jump, tension can

So, quick revision. One: a spring force cannot jump; its extension needs time to change. Two: tension and normal force can jump, and a cut string pulls zero at once. Three: just after the cut, keep the spring forces, remove what was cut, and find each acceleration from what remains. Now go and try the practice problems on this page. All the best!