Thrust: Forces from a Stream of Mass

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Picture a rocket far out in space. No air around it, no ground below it, nothing at all to push against. The engine fires, and the rocket speeds up. But it pushes against what? Strange, right? Today's idea, thrust, answers this. And it comes up often in JEE problems, from guns to rockets. So, let's begin.

0:29Momentum flow: one formula for every stream

So, here is the set-up. Some mass is being thrown out, or brought to rest, at a steady rate. That rate is d m over d t, the fraction on screen: a tiny mass d m in a tiny time d t, so kilograms per second. Steady means this rate does not change. Okay? Now, each kilogram moves at a speed u relative to whatever throws it out or stops it. So tell me, how much momentum does this stream carry every second? u times d m over d t, yes, exactly! Every second, d m over d t kilograms go by, and each kilogram carries momentum u. And remember, u is the relative speed. Clear? Now, the second law. Force is the rate of change of momentum, right? So momentum per second is a force: F equals u times d m over d t. This one line is the heart of today's lesson. Every case that follows is just this formula in a different costume. Got it?

1:28Machine gun and liquid jet

First case, a machine gun. It fires n bullets every second, each of mass m, at speed v. So, what is d m over d t here? n times m, correct! Every second, n bullets, each of mass m. And the relative speed u is just v. The double arrow means implies, so F equals n m v. This is the force needed to hold the gun steady, okay? Next, a jet of liquid. Rho, the Greek letter that looks like a p, is the density; capital A is the area of the jet, and v is its speed. In one second the jet moves forward v metres, so a column of volume A times v goes by, with mass rho A v. With me? Now the wall stops the liquid, so the relative speed is v again. F equals rho A v, the mass per second in brackets, times v. So F equals rho A v squared; the small two on top means squared. Right? Now, what if the liquid does not stop, but bounces back with the same speed? More force, or less? More, double in fact! Each kilogram goes from v forwards to v backwards, a change of two v. So F equals two rho A v squared. Just like the ball rebounding from a wall in impulse: the speeds add. But only when it rebounds with the same speed, okay?

2:53Rocket thrust

Now, our puzzle: the rocket. The engine throws gas out of the back, downwards, at speed u relative to the rocket. Every second, the gas carries away momentum u times d m over d t, downwards. So the rocket gets that same force, upwards. This is the thrust. See, the rocket does not need air to push on. It pushes on its own gas, and the gas pushes back! Puzzle solved. But near the Earth, is the thrust the only force on the rocket? Think. Its weight, m g, yes! Mass m times g, the acceleration due to gravity, pulling down. So the net force, m a, equals the thrust minus m g. Here a is the rocket's acceleration, and m is its mass at that moment. Write this down; we need it in example two. Right? And far out in space, with no gravity, a rocket starting from rest reaches a speed v equal to u times l n of m nought over m. Here l n is the natural logarithm, m nought, the m with a small zero below, is the starting mass, and m is the mass left. So for a given u, what matters is how many times the mass has shrunk.

4:06Conveyor belt: bringing mass up to speed

Clear? Last case, a conveyor belt moving at a steady speed v. Sand is dropped on it gently, so it lands with no forward speed. The speed is steady, so why does the belt need any extra force at all? Because every bit of sand lands at rest and must be brought up to speed v, right? Each second, d m over d t kilograms of sand gain speed v. So the extra force is v times d m over d t: the same formula, with relative speed v. And note the conditions: dropped gently, belt at a steady speed. Got it?

4:42Example 1: force needed to hold a machine gun

Alright, example one. Here, a machine gun fires ten bullets per second, each of twenty grams, at five hundred metres per second. On screen, g is grams, and m s to the minus one is metres per second. Find the force needed to hold it. Pause the video, and try it first. Done? Okay, let's check. Many students write ten times twenty times five hundred. One hundred thousand newtons! Is that right? Oh no, careful! The mass is in grams. Change it to kilograms first: twenty grams is zero point zero two kilogram, and k g on screen is kilogram. Okay? Now F equals n m v: ten times zero point zero two times five hundred. One hundred newtons; capital N here is newton, the unit. That is the push needed to hold the gun steady. Got it?

5:40Example 2: fuel burn rate for a rocket to hover and rise

Now a slightly harder one, and our rocket is back! Here, a five thousand kilogram rocket ejects gas at eight hundred metres per second relative to itself. At what rate must it burn fuel, part a, just to hover, and part b, to rise with an acceleration of ten metres per second squared? That is the m s to the minus two on screen. And take g as ten, as the badge shows. Pause and try. Done? Okay, let's check. Part a, hover: the rocket just hangs in the air, with no acceleration. So the thrust equals the weight: u times d m over d t equals m g, five thousand times ten. That is five times ten to the four newtons; the small four on top is a power of ten, so fifty thousand newtons. With me? So d m over d t is fifty thousand over eight hundred: sixty two point five kilograms per second. On screen, k g s to the minus one means kilograms per second. Right? Now part b. Many students say: thrust equals m a, five thousand times ten, so the same sixty two point five again. Correct? No, no, wait! This is the trap. Gravity has not switched off. The net force is the thrust minus m g, and that equals m a. So the thrust equals m times, g plus a. Clear? So the thrust is five thousand times, ten plus ten, twenty: ten to the five newtons, one hundred thousand newtons. Okay? And d m over d t is ten to the five over eight hundred: one hundred and twenty five kilograms per second. Double the hover rate! Forget m g, and you get only half the answer. Got it?

7:28Quick revision: F = u dm/dt for guns, jets and rockets

So, quick revision. One: for a stream of mass, F equals u times d m over d t, the momentum carried per second. Two: for a gun, F equals n m v; for a jet, F equals rho A v squared, doubled if it rebounds with the same speed. Three: for a rocket, m a equals the thrust minus m g; never forget the weight. Now go and try the practice problems on this page. All the best!