Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a block starting from rest at a height of three R on a smooth track, with no friction at all. At the bottom, it runs into a vertical loop of radius R, okay? When it reaches the side of the loop, where the radius is horizontal, how hard does the track push on it? Just its own weight, m g? Maybe a little more? Hmm, hold on to your answer. You have finished Chapters one to eight of work, energy and power, 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 we will solve two of its problems together. So, let's begin.

0:43How to use this chapter

So, let's open this chapter and see how to use it. Part A has eighteen solved problems, at JEE Main level. Okay? And the course is very clear: attempt Part A as a forty minute test, after you finish Chapters one to eight. So first the chapters, then sit down with a clock, just like a real test. Right? Now, a quick one for you. Eighteen problems, forty minutes. How much time can you give each problem? Forty over eighteen; that is the division sign on screen. About two point two, and the wavy equals sign means approximately. So, a little over two minutes a problem, yes! Read it, spot the idea, solve it, and move on. Clear? And one rule for the whole chapter: take g, the acceleration due to gravity, as ten. On screen, m s to the power minus two means metres per second squared. Got it?

1:39Part B: two or three ideas in one problem

Now Part B: twelve longer problems, at JEE Advanced level. With me? But tell me, what makes a problem Advanced? Bigger numbers? Longer calculations? Not really! These problems join two or three ideas of the unit. Each idea alone is easy for you now. The real skill is spotting where one idea ends and the next one begins. Okay? For example, a block slides down an incline with friction, and then meets a spring. Friction on an incline, plus a spring; the plus sign joins the two ideas. Right? Then, a bullet passes horizontally through a pendulum bob. First momentum, across the impact, and then a vertical circle for the bob. Clear? And a block on a smooth track: energy conservation, plus circular motion. Remember our puzzle? That is exactly this problem, and it will be our Example two. Got it?

2:35Part C and the unit test

Now for Part C: fifteen short concept checks, on the ideas that are often tested as statements. Okay? Let me give you one. Is the work done by static friction always zero? Static friction means no sliding, so no work, right? No, no, careful! Think of a box on the floor of a truck that is speeding up. Static friction drags the box along with the truck, so in the ground frame, it does positive work on the box. One word, always, made the statement false. So read every word slowly. With me? And at the very end of the chapter, the unit test: fourteen questions, a quick recall drill. Answer them fast, from memory; that is what a recall drill means. Clear?

3:20Example 1: work of a constant force along a curve

Alright, here is an easy one from Part A to start with. Read it with me: a particle moves from the point zero, zero to the point three, two, in metres, along the curve y equals two ninths x squared. The force on it is constant: F equals two i hat plus four j hat, in newtons. The little arrow on F means it is a vector, and i hat and j hat, with their little caps, are the unit vectors along x and y. The capital N is the newton. Find the work done. Pause the video, and try it. Done? Okay, let's check. The path is a curve. So we must integrate the force along the whole curve, bit by bit, right? No, no, wait! Look again: the force is constant. For a constant force, only the displacement matters, from start to end, not the path in between. So forget the curve completely. Okay? The displacement, delta r, with the little triangle for delta, is the final point minus the starting point: three i hat plus two j hat, in metres. Right? Now the work, W, is F dot delta r; the dot means the dot product. Multiply the matching parts and add: two times three, plus four times two. Clear? Six plus eight: fourteen joules; the capital J is the joule. One line, well inside your two minutes. Very good!

4:50Example 2: track force at the side of a vertical loop

Now our puzzle from the start, as a Part B problem. Read it with me: a block is released from rest at a height of three R, on a smooth track, and enters a vertical loop of radius R, okay? Find the track force on it where the radius of the loop is horizontal. Pause the video, and try it. Done? Okay, let's check. First, the figure. The dot marked start is the release point, at height h; for us, h is three R, okay? The arrow marked R is the radius, drawn horizontally, and its right end is exactly our point on the side of the loop. The labels at the top and on the right, v squared greater than or equal to g R, and h min, that is h with a small min below, equals five over two R, are the conditions for just getting round. Our three R is more than enough. Clear? Now the speed there. The block started at three R, so it has dropped through three R, right? Oh no, careful! At the side of the loop, the block is still R above the bottom, not at the bottom. So the drop is three R minus R: only two R, got it? The track is smooth, so there is no friction, and mechanical energy is conserved. It started from rest, so v squared equals two g, times, in brackets, three R minus R, which gives four g R, with me? Now the second idea: circular motion. Two forces act on the block here: its weight, m g, downwards, and the track force, capital N, pushing from the track. Which of them points towards the centre? Only N, exactly! Here the radius is horizontal, so the track runs straight up and down, and the weight, m g, acts along the track, not towards the centre. So the weight gives no centripetal push at all. Right? So the track force alone is centripetal: N equals m v squared over R, got it? Now we put four g R in place of v squared. N equals m times four g R, over R; the dot here just means times. The R cancels, and N is four m g. Four times its own weight! So in our puzzle, just m g was far too small. Puzzle settled. Energy first, then circular motion: two ideas in one problem, exactly the Part B style. Perfect!

7:12Quick revision: test plan, constant-force work, loop force

So, quick revision. Part A is a forty minute test, Part B joins two or three ideas in one problem, and g is ten. For a constant force, the work is F dot delta r, whatever the path. And on a loop, where the radius is horizontal, only the track force N is centripetal: N equals m v squared over R, with no m g in it. Now open Part A, start your forty minute clock, and begin. Then Parts B and C, and the unit test at the end. You did the hard work in eight chapters, and this is where it pays off. All the best!