Part B: Harder Problems

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take two identical balls, each going round a vertical circle of the same radius. One is tied to a string. The other runs inside a smooth, hollow tube. At the lowest point, both have exactly the same speed, root four g l. One of them goes all the way round, and the other one does not. Which one? Hmm, same speed, same circle, so how can the answers be different, right? Today is Part B of the Work, Energy and Power problem bank: the harder problems, where two or three ideas meet in one question. Problems like these come up often in JEE Advanced, so let's begin.

0:45Idea 1: a string only pulls, a tube pushes both ways

So, let's settle the puzzle. Remember, a string can only pull; it can never push. Now tell me, which ball has the harder job: the one on the string, or the one in the tube? The string, yes! Here, u is the speed at the lowest point, and capital T is the tension in the string. At the top, the string must stay tight, so T must be greater than or equal to zero; that is this sign. That needs v top, the speed at the top with the word top written below, to be at least root g l, the square root of g times l. Here g is the acceleration due to gravity, and l is the length of the string, the radius. With no friction, mechanical energy is conserved from the bottom to the top, and that gives: u at least root five g l. Clear? Now the tube. The tube can push from outside and from inside, both ways. So the ball only has to reach the top: v top at least zero is enough. The tube is smooth, so energy conservation then gives u at least root four g l, which is two root g l. So, our puzzle: with root four g l, the ball in the tube just makes it round, while the string needs root five g l, so that string goes slack before the top.

1:57Idea 2: too slow for the top of a tube

Got it? Now let's use this. Read it with me: a particle in a smooth vertical circular tube of radius l has speed root three g l at the lowest point. Does it reach the top? If not, how high does it rise? First part: does it reach the top? No, it does not! Even in a tube, it needs root four g l, and it has only root three g l. Okay? So where does it stop? The tube is smooth, no friction, so mechanical energy is conserved. The kinetic energy at the bottom, half m times three g l, all becomes potential energy, m g h. Here m is the mass, and h is the height above the lowest point. Cancel m and g, and h equals one point five l. Right? Now look carefully. One point five l is more than l, so this point is above the centre of the circle. Up there, a slow body on a string would go slack and leave the circle. So does our particle fall off the circle too? No, no, it does not! That is the trap. Yes, one point five l is greater than l; this sign means greater than. But a tube is not a string. Up there, gravity pulls the slow particle inwards more than it needs, so the inner wall of the tube pushes it outwards, and holds it on the circle. But a string can only pull inwards, so it could never do that. Clear? So the particle simply stops at a height of one point five l, and slides back down. String and tube, same speed, different answers. Always check which one the question gives you!

3:32Example 1: power of a pump that lifts and ejects water

Now, example one, a medium one. Read it with me: a pump raises water through five metres, and ejects it at twenty metres per second; the m s to the power minus one means metres per second. It comes out through a nozzle of area four square centimetres; the c m with a little two is square centimetres. Find its power. Pause the video, and try it first. Done? Okay, let's check. First, how much water comes out every second? m dot, the m with a dot on top, is the mass per second. It equals rho, this Greek letter, the density of water, one thousand kilograms per cubic metre, times the area A, times the speed v. Four square centimetres is four times ten to the power minus four square metres. So one thousand, times four times ten to the power minus four, times twenty: eight kilograms per second. With me? Now the energy given to each kilogram. The pump lifts it through five metres, so it gains g h: ten times five, fifty joules of potential energy. Here g is ten metres per second squared, and the capital J is joules. Okay? So the power is just m dot times g h. Eight times fifty, four hundred watts, right? Oh no, careful! That is the trap. The water does not just rise; it also leaves the nozzle at twenty metres per second. So each kilogram also gets half v squared: half times twenty squared, two hundred joules of kinetic energy. That is four times the potential energy! See? So the power, capital P, equals m dot, times g h plus half v squared. Eight times, fifty plus two hundred: two thousand watts, where the capital W is the watt. That is two kilowatts; k W means kilowatts. Forget the kinetic energy, and you lose four fifths of the answer! Got it?

5:31Example 2: bullet through a bob that just goes round

Now example two, a harder one, JEE Advanced style. Read it with me: a bullet of mass small m passes horizontally through the bob, of mass capital M, of a pendulum of length l, and emerges with half its speed. Find the least initial speed for which the bob completes a vertical circle. Pause the video and try it. Hint: first ask what speed the bob needs. Done? Okay, let's check. Start with the bob. Just after the bullet passes, what speed must the bob have at the bottom: root four g l, or root five g l? Root five g l, yes! This bob hangs on a string; it is not inside a tube. The string must stay tight at the top, so the bob needs root five g l at the bottom. That is exactly our Idea one. Clear? Now the bullet. It goes in with speed u, and comes out with half its speed. Can we write conservation of kinetic energy for the bullet and the bob? No, no, careful! The bullet tears through the bob, and a lot of energy turns into heat. But the impact is very short, and at that moment the string pulls only vertically. So horizontal momentum is conserved. Okay? So, momentum before equals momentum after. m u equals m times u over two, for the bullet, plus capital M times root five g l, for the bob. With me? Subtract m times u over two from both sides: m times u over two equals capital M root five g l. So u equals two capital M over small m, times root five g l. Momentum for the impact, and the vertical circle for the bob: two ideas in one question. That's Part B!

7:16Quick revision: string vs tube, pump power, bullet

So, quick revision. To go round a vertical circle, a ball on a string needs root five g l at the bottom, but a ball in a tube needs only root four g l, because the tube pushes both ways. For a pump, the power is m dot, times g h plus half v squared: don't forget the kinetic energy. And when a bullet passes through a bob, momentum is conserved, but kinetic energy is not. Now try the practice problems on this page. Take your time; these are the harder ones. All the best!