Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Let go of a five kilogram block, and it falls with g, ten metres per second squared, right? Now wind a string on a ring-shaped pulley, also of five kilograms, turning on a fixed axle, and hang the same block from it. Let go again. This time the block falls at exactly half of g. Why half? Where did the rest go? Hold on to that question. Today: how this chapter of mixed problems works, the five big ideas of rotational motion that come up often in JEE Main, and two solved problems, including this exact puzzle. So, let's begin.

0:41How this chapter works

So, let's open this chapter. Recent JEE Main questions on rotational motion test five main ideas. Okay? And where are the real past questions, from twenty twenty one to twenty twenty six? They sit at the end of Chapters one to five. Right? This chapter is different. Its problems are original, written in the same style, so you can practise the pattern. With me? And one rule for the whole chapter. g, the acceleration due to gravity, is ten metres per second squared, unless a question says otherwise. That is what m s to the power minus two means on screen. Clear? There are three sets of eight solved problems, and each set ends with five practice problems. Got it?

1:26Ideas 1 and 2: centre of mass, parallel axes

Now, the five ideas. Idea one: the centre of mass of a composite body, a body made of simple parts. The trick: replace each part by a point mass, placed at that part's own centre of mass. Okay? Then use the formula for particles. r c m, with the small arrow on top for a vector, equals the sum of m i times r i, over capital M, the total mass. The big Greek sigma means add them all up, and the small i labels each part. With me? Idea two: moments of inertia by the parallel axis theorem. I equals ICM plus capital M d squared. ICM is the moment of inertia about an axis through the centre of mass, and d is the distance to the new, parallel axis. Now tell me: can you start from any axis you like, and just add capital M d squared? No, no, careful! The theorem always starts from the axis through the centre of mass. Starting from some other axis is a common trap. Right?

2:28Ideas 3 to 5: pulleys, angular momentum, rolling

Idea three: tau equals I alpha, with pulleys. Tau, the Greek letter, is the torque, I is the moment of inertia, and alpha is the angular acceleration. This holds for a rigid body about a fixed axis, or an axis through the centre of mass. If the string does not slip, a equals alpha R, with R the radius. And with a heavy pulley, the tensions on the two sides are different. Clear? Idea four: conservation of angular momentum. If the net external torque about an axis is zero, then I one omega one equals I two omega two, about that same axis. Omega, the Greek letter, is the angular speed, and the small one and two mean before and after. Okay? Idea five: rolling down an incline, without slipping. a equals g sine theta, over one plus k squared over R squared. Theta is the angle of the incline, and k is the radius of gyration. Now, a quick race. Take a heavy ring and a light disc, rolling from rest down the same incline. Which one reaches the bottom first? The disc, yes! For a disc, k squared over R squared is half; for a ring, it is one. Smaller k squared over R squared, bigger acceleration. Mass and radius do not matter at all, only the shape. Got it?

3:49Example 1: block on a ring pulley falls at half g

Now, Example one, from Set one of this chapter. Here, a five kilogram block hangs from a string wound on a ring-shaped pulley of mass five kilograms, with a fixed axle. Find the acceleration and the tension. Yes, this is our puzzle. Pause the video, and try it first. Done? Okay, let's check. Small m is the mass of the block. The course gives this result for a block on a drum: a equals m g, over m plus I over R squared. Our pulley is a ring. So what is I over R squared? For a ring about its axis, I equals capital M R squared, where capital M is the mass of the pulley. The double arrow on screen means 'so', and I over R squared is just capital M, five kilograms. Right? So a equals m g over m plus capital M, as on screen. Five times ten, the cross meaning times, over five plus five: five metres per second squared. Half of g! There is our puzzle, solved: the ring acts like five extra kilograms the block must drag along. With me? Now the tension. Many students write: tension equals m g, fifty newtons. Correct? Oh no, careful! The block is accelerating down, so m g minus T equals m a. T equals m times g minus a, in brackets: five times ten minus five, which is twenty five newtons. Clear? And a quick check with the pulley. T times R equals I alpha, and for our ring this simplifies to T equals capital M a. Five times five: twenty five newtons again. Perfect!

5:35Example 2: particle sticks to a pivoted rod, find omega

Now a harder one, from Set two. Here, a particle of mass m, moving with speed v on a smooth table, strikes the free end of a uniform rod at right angles, and sticks. The rod has mass two m and length capital L, and it is pivoted at its other end on the table. Find omega, the angular speed just after. Pause and try. Hint: look at the pivot. Done? Okay, first things first. Momentum before equals momentum after. Can we write that here? No, we cannot! The pivot holds the end of the rod in place, so during the impact it pushes back with a big force. So linear momentum is not conserved. But that force acts right at the pivot, so its torque about the pivot is zero. So, conserve angular momentum about the pivot. Got it? Before the impact, only the particle moves. It moves at right angles to the rod, at the free end, a distance L from the pivot. So its angular momentum is m v times L, okay? Just after, the rod and the particle turn together. Now, the rod's moment of inertia: one twelfth of two m L squared, or one third? One third, because the rod turns about its end, not its centre. So one third of two m L squared. Plus the particle, now stuck at distance L: m L squared. Do not forget this part! Total: five thirds m L squared. Right? So, m v L equals five thirds m L squared times omega. Cancel the m, and one power of L, from both sides. Omega equals three v over five L, that's it! One law, about the right point, and the problem is done. With me?

7:24Quick revision: five ideas, drum formula, pivot collisions

So, quick revision. Five ideas: centre of mass, parallel axes, pulleys, angular momentum, and rolling. For a block hanging from a string wound on a pulley or drum, a equals m g over m plus I over R squared, less than g. And when a pivot is involved in a collision, conserve angular momentum about the pivot. Now go and try the practice problems in the three sets of this chapter. All the best!