Work, Energy and Power in Rotation

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Think of a heavy flywheel, the spinning wheel inside an engine. From rest, you spin it up to six hundred revolutions per minute. Now you want twelve hundred, double the speed. For the next six hundred, will you need the same work again? The same? Twice? More? Keep your answer in mind; we will calculate it. Today: work, energy and power in rotation. This comes up often in JEE, so let's begin.

0:33Work and power in rotation

So, start from what you know. In a straight line, work W equals force F times distance s. And power P equals F times v, force times velocity. Okay? Now think. When a body turns, what plays the role of force? And what plays the role of distance? Torque and angle, yes, exactly! So, in rotation, W equals the integral of tau d theta. Tau, this Greek letter, is the torque, and theta, the other one, is the angle turned. The tall S-shaped sign is the integral sign: it adds up tau times every tiny angle d theta. With me? Often the torque stays constant. Then the integral is simply tau times theta. But careful: theta must be in radians, not revolutions, not degrees. One revolution, rev for short, is two pi radians, okay? Next, power: the work done per second. If W is tau times theta, what is P? Tau times omega, correct! Theta per second is the angular speed, omega, this Greek letter. So P equals tau omega, with omega in radians per second. See? Torque replaces force, and omega replaces velocity. Clear?

1:55Work-energy theorem in rotation

Now, energy. For a rigid body turning about a fixed axis, the net work done on it equals the change in half I omega squared. Capital I is the moment of inertia, the small net below W means the work of all the torques together, and this little triangle, delta, means change in. Okay? Change means final minus initial: half I omega two squared, minus half I omega one squared. The small one and two below omega mean start and end. We need this for our flywheel puzzle. Right? Now a different case. Take a wheel rolling on a road. It moves forward and spins at the same time. What is its kinetic energy? Both, added! Yes. K equals half capital M, v c m squared, plus half I c m, omega squared. Capital M is the mass, and the small c m below means centre of mass. One term for moving, one for spinning. Students often write only one of them. Don't! Got it?

2:59Energy with a pulley

Now, pulleys: the figure from our last lesson. Blocks m one and m two hang over a pulley of mass capital M and radius capital R, okay? The heavy block, m one, goes down with acceleration a; the pulley turns with angular acceleration alpha. T one and T two are the tensions, and the equations on the right are last lesson's force method. Today, we use energy instead. With me? So, the heavy block falls and the light one rises. Overall, the blocks lose potential energy. Where does that energy go? Into kinetic energy! But not only of the blocks. The pulley spins too, so it takes a share. So, the P E lost by the blocks equals the K E gained by the blocks plus the K E gained by the pulley. P E and K E mean potential and kinetic energy. Okay? Now, the link. The string does not slip, so the rim moves with the string: v, the speed of the blocks, equals omega times capital R, the radius. Right? So the pulley's kinetic energy, half I omega squared, becomes half, times I over R squared in brackets, times v squared. Exactly like one more block, of mass I over R squared! That's the topper's trick: a pulley simply adds an extra mass, I over R squared, to the blocks. Clear? Keep two values ready. For a disc, I over R squared is half of capital M, okay? For a ring, all the mass sits on the rim, so it is the full mass, capital M, right?

4:35Example 1: flywheel speeded up from 600 to 1200 rpm

Alright, Example one: our flywheel puzzle. Here, a flywheel with I equals zero point five kilogram metre squared is speeded up from six hundred to twelve hundred r p m. Find the work done. And r p m means revolutions per minute. Pause the video, and try it first. Done? Okay, let's check. The speeds are given in r p m. Can we put six hundred and twelve hundred straight into the formula? No, no! Omega must be in radians per second. Each revolution is two pi radians, and a minute is sixty seconds. So six hundred r p m is six hundred times two pi over sixty: twenty pi radians per second. The s to the power minus one means per second. Okay? In the same way, twelve hundred r p m is forty pi radians per second. Right? Now the work. Many students subtract the speeds first: forty pi minus twenty pi, then half I times that difference squared. Is that correct? Oh no, careful! That is the trap. The theorem says change in half I omega squared: square each omega first, then subtract. So W equals half I, times omega two squared minus omega one squared: half times zero point five, times sixteen hundred pi squared minus four hundred pi squared. With me? That is three hundred pi squared: about two thousand nine hundred and sixty one joules, or about two point nine six kilojoules. The wavy equals sign means approximately, okay? Now, our puzzle. At six hundred r p m, K one is half times zero point five, times four hundred pi squared: one hundred pi squared joules. That was all the work from rest. But the next six hundred r p m cost three hundred pi squared. So W is three times K one! Because energy goes as omega squared. Got it?

6:37Example 2: Atwood machine with a ring pulley, by energy

Now a harder one, JEE level. Blocks of three kilograms and one kilogram hang over a ring-shaped pulley of mass two kilograms. Using energy, find their speed after the heavier block has fallen one metre from rest. Take g as ten metres per second squared. Pause and try. Hint: the pulley is just extra mass. Done? Okay, let's check. The pulley is a ring, so I over R squared is the full mass, capital M: two kilograms. Got it? Now, the potential energy lost. The three kilogram block falls one metre: three times ten times one, thirty joules. So thirty joules lost, right? No, no, wait! As the heavy block falls one metre, the one kilogram block rises one metre and gains ten joules. So the net loss is three minus one, times g, times one: twenty joules. Clear? Now, the kinetic energy gained. Both blocks move with the same speed v, and the ring adds two kilograms of extra mass. So half, times three plus one plus two, times v squared: that is three v squared. With me? So twenty equals three v squared. The double arrow means 'which gives': v equals the square root, the sign on screen, of twenty over three: about two point five eight metres per second. See the two in that bracket? That is the pulley's share. No tensions, no torques: just energy!

8:11Quick revision: work, power and energy with pulleys

So, quick revision. W equals tau theta, with theta in radians, and P equals tau omega. The net work is the change in half I omega squared: square each omega first, then subtract. And for pulleys, the potential energy lost goes into kinetic energy of the blocks and the pulley, with I over R squared as extra mass. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!