Newton's Second Law for Rotation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Two blocks hang over a pulley: five kilograms on one side, and three on the other. In the laws of motion chapter, the pulley was light, and the tension was the same on both sides, right? Now give the pulley some mass. If both sides still pull with the same tension, their turning effects cancel. So how does the pulley ever start spinning? Something has to change. Today we find out what. The answer is Newton's second law for rotation, and pulley problems built on it come up often in JEE, so let's begin.

0:40Newton's second law for rotation

So, let's start from what you already know. For translation, Newton's second law says F equals m a: the net force F equals the mass m times the acceleration a. Okay? Now think. In rotation, what makes a body spin faster and faster? What plays the role of force? Torque, yes, exactly! So, for rotation: tau ext equals I alpha. Tau, the Greek letter, is the torque, and the small e x t below it means external: only torques from outside the body count. Capital I is the moment of inertia, and alpha, the other Greek letter, is the angular acceleration. Clear? Now compare the two lines. If torque is like force, and alpha is like acceleration, then what is like mass? The moment of inertia, correct! The double-headed arrows on screen mean: plays the role of. Torque for force, I for mass, alpha for acceleration. So a large moment of inertia resists a change in spin, just as a large mass resists a push. Got it? But this law needs a condition. It holds for a rigid body rotating about a fixed axis, or about an axis through the centre of mass. And take the torque and the moment of inertia about that same axis.

2:00Strings on pulleys

Right? Now, pulleys. Look at this figure. Here is a pulley of mass capital M and radius capital R, hanging from the ceiling. Over it runs a string with two blocks, m one and m two: the small one and two below are just labels. m one is heavier, so it goes down with acceleration a, the green arrow, and the pulley turns with angular acceleration alpha, the curved green arrow at the top. On the right is the full working, and we will build it line by line. With me? First, the string does not slip on the pulley. So the rim moves along with the string, and a equals alpha R, okay? Now our puzzle. T one is the tension on the heavy side, and T two is the tension on the light side. Can they be equal now, with a heavy pulley? No, no, they cannot! T one and T two pull the rim in opposite senses, each with lever arm R, the radius. If they were equal, the net torque would be zero, and the pulley could never speed up. So, T one minus T two, in brackets, times R, equals I alpha. Taking one tension T on both sides of a heavy pulley is the most common trap here. Don't fall for it! Clear? Now add Newton's law for each block: the first two lines on the right of the figure. Put them together, and you get: a equals m one minus m two, times g, over m one plus m two plus I over R squared. Here g is the acceleration due to gravity. Look at the bottom. I over R squared sits next to the masses, like extra mass the blocks must drag along. Got it?

3:40One block on a drum

Now a single block of mass m hangs from a string wound on a drum. For the block, weight down and tension up: m g minus T equals m a. Right? For the drum, with moment of inertia I and radius R, the tension gives the torque: T times R equals I alpha. The string does not slip, so alpha is a over R, okay? The double arrow means so: T equals I over R squared, times a. With me? Now add the two equations. T cancels, and m g equals m plus I over R squared, in brackets, times a. So, a equals m g over m plus I over R squared. Write this one down. Clear? Now tell me. Does this block fall freely, with acceleration g? No, slower than g! The bottom is bigger than m, so a is less than g. That sign on screen is the less than sign. Once again, the drum's I over R squared acts like extra mass. With me?

4:42Example 1: angular acceleration of a disc on an axle

Alright, Example one. Here, a ten newton force is applied tangentially to the rim of a uniform disc. The disc has mass two kilograms and radius zero point five metres, and it is mounted on a fixed axle. Find alpha, its angular acceleration. Pause the video, and try it first. Done? Okay, let's check. The disc turns on a fixed axle, so tau equals I alpha applies directly. Right? First, the moment of inertia. For a uniform disc about its own axis, I equals half M R squared. Half, times two, times zero point five squared: zero point two five kilogram metre squared. Okay? Now the torque. The force acts along the rim, tangentially. So what is its lever arm about the axle? The radius itself, exactly! So tau equals F times R: ten times zero point five, which is five newton metre. With me? So alpha equals tau over I: five over zero point two five, which is twenty radians per second squared. That is what s to the power minus two means on screen. And careful: the radius is squared in I, but not in the torque.

6:02Example 2: Atwood machine with a heavy disc pulley

Got it? Now a harder one, our puzzle from the start. Blocks of five kilograms and three kilograms hang over a disc-shaped pulley of mass four kilograms. Find the acceleration and the two tensions. Take g as ten metres per second squared. Pause and try. Hint: the two tensions are not equal. Done? Okay, let's check. We use the formula from the figure. But first, the pulley is a disc. So what is I over R squared? For a disc, I is half M R squared, so I over R squared is half M: half of four, that is two kilograms. The pulley acts like two extra kilograms to drag along. Right? Now the acceleration: five minus three, times g, over five plus three plus half of four. With g as ten, the top is twenty and the bottom is ten. Twenty over ten: two metres per second squared. Clear? The five kilogram block goes down, so T one equals five times ten minus two, in brackets: forty newtons. Okay? Now the other side. Many students just write forty newtons again. Is that correct? No, careful! The three kilogram block goes up, so T two equals three times ten plus two: thirty six newtons. So the two tensions really are different. With me? Check with the pulley: T one minus T two should equal half M a. Half times four times two: four newtons. And forty minus thirty six is four. Perfect! There is our puzzle, solved: this four newton difference gives the torque that spins the pulley. Got it?

7:47Quick revision: torque = I alpha, a = alpha R, block on drum

So, quick revision. Tau ext equals I alpha, for a rigid body about a fixed axis, or an axis through the centre of mass. When the string does not slip, a equals alpha R, and when a heavy pulley has angular acceleration, the tensions on its two sides are different. And a block on a drum falls with a equals m g over m plus I over R squared, less than g. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!