Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a heavy disc, spinning freely about its axle. Now you gently drop a second disc onto it, at rest, on the same axis. Soon the two discs spin together, a little slower. Angular momentum about the axis is conserved here. So the kinetic energy is conserved too, right? Hmm, are you sure? Hold that thought. We will settle it in Example two, with real numbers. Rotational motion comes up often in JEE, and this chapter, the unit problem bank, is where you test yourself on the whole unit. Today I will show you how to use it, and then we will solve two of its problems together. So, let's begin.

0:46Part A: a timed test

So, let's open this chapter and see how to use it. First, Part A: eight solved problems, mixed from the whole unit. Okay? And the course is very clear: attempt Part A as a thirty minute test, after you finish Chapters one to six. So first the chapters, then sit down with a clock, exactly like a real test. Right? Now, a quick one for you. Eight problems, thirty minutes. How much time can you give each problem? Thirty over eight; that is the division sign on screen. Three point seven five. So a little under four minutes a problem, yes, exactly! Read it, spot the idea, solve it, and move on. Clear?

1:30Part B: concept checks

Now, Part B: six concept checks. Here you test the ideas themselves. With me? And why these six? Because these concepts are often asked as statement or assertion questions. Right? Let me give you one from Part B, okay? The assertion: a diver increases her rate of spin by pulling in her arms and legs. Clear? And the reason: when she tucks in, the torque on her increases. So tell me. Is the assertion true? And is the reason true? The assertion is true, yes! She really does spin faster. But the reason? Oh no, careful! It is false. Gravity acts at her centre of mass, so about her centre of mass, gravity gives no torque at all. Got it? No torque, so her angular momentum, capital L, stays constant. And L equals I times omega: capital I is her moment of inertia, and omega, the Greek letter, is her angular speed. When she tucks in, I falls, so omega rises. See, a true assertion can come with a false reason. So judge the two parts separately. Clear?

2:40Part C: JEE Advanced level

And finally, Part C: four longer problems, at JEE Advanced level. Okay? What kind of problems? The force on the hinge of a rod falling from the horizontal, and a sphere rolling without slipping on a plank that is being pulled. Right? Then an impulse at one end of a rod lying free on a smooth table, and a bead set moving on a ring that is free to turn. Each one needs two or three ideas of the unit at the same time. With me?

3:09Example 1: torque and work done on a wheel

Alright, here is an easy one from Part A to start with. Read it with me: a wheel has moment of inertia, capital I, equal to four kilogram metre squared. Torque increases its angular speed from ten to thirty radians per second, in five seconds; the s to the power minus one means per second. Find the torque and the work done. Pause the video, and try it. Done? Okay, let's check. Torque first. For rotation about a fixed axis, tau, the Greek letter, equals I times alpha, the angular acceleration. And alpha is capital delta omega over capital delta t; the little triangle, capital delta, means change in. Okay? So tau equals four times, thirty minus ten, over five. Four times twenty is eighty, and eighty over five is sixteen. Sixteen newton metre; that is the capital N and small m on screen. Very good! Now the work. The change in omega is twenty. So the work, capital W, is half times four times twenty squared, eight hundred joules, right? No, no, careful! That is the trap. The work done by the torque equals the change in kinetic energy, capital delta K, okay? Kinetic energy is half I omega squared, so we need half I times, omega two squared minus omega one squared. The small two and one below omega mean final and initial. Not the change in omega, squared! Clear? Thirty squared is nine hundred, and ten squared is one hundred. Half times four times eight hundred: sixteen hundred joules; capital J is the joule. Double the trap answer! See why the right formula matters?

4:56Example 2: disc dropped on spinning disc, heat produced

Now our puzzle from the start, as a Part A problem. Read it with me: a horizontal disc, with I equal to zero point five kilogram metre squared, spins freely at twelve radians per second. Then a second disc, with I equal to zero point two five kilogram metre squared, at rest, is dropped onto it coaxially, that means on the same axis. Find the common angular speed, and the heat produced. Pause the video, and try it. Done? Okay, let's check. First, our puzzle. I told you angular momentum about the axis is conserved here. But why? Because there is no external torque about the axis, yes, exactly! The first disc spins freely, and the forces between the two discs act inside the system. So the angular momentum, capital L, about that axis stays the same. Okay? Before: only the first disc turns, so L is I one times omega one; the small one below means the first disc. After: both discs turn together at one common omega, so L is, I one plus I two, times omega. Set them equal. Right? So omega equals zero point five times twelve, over zero point seven five; that is I one plus I two, zero point five plus zero point two five. Zero point five times twelve is six, and six over zero point seven five is eight. Eight radians per second. Perfect! Now the energy. Before, K one is half times zero point five times twelve squared. Twelve squared is one hundred and forty four, so K one is thirty six joules. With me? After, K two is half times zero point seven five times eight squared. Eight squared is sixty four, so K two is twenty four joules. Got it? So the kinetic energy is not conserved! Thirty six minus twenty four: twelve joules turned into heat, as the discs rubbed until they moved together. That is one third of the starting energy. So, our puzzle: angular momentum is conserved, but kinetic energy is not. Conserve energy here, and your omega comes out wrong. Very good!

7:14Quick revision: work is change in KE; L conserved, KE not

So, quick revision. Part A: eight problems as a thirty minute test, after Chapters one to six. Part B: statement or assertion questions, so judge each part separately. Part C: longer problems at JEE Advanced level. Work done equals the change in kinetic energy. And with no external torque, angular momentum is conserved, but kinetic energy need not be. Now set your thirty minute timer, start Part A, and go. Then try the practice problems on each page. You can do this, I promise. All the best!