Conservation of Mechanical Energy

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. From a cliff twenty five metres high, you throw three stones, each at ten metres per second: one horizontally, one at thirty degrees above the horizontal, and one straight down. Which stone hits the ground fastest? Straight down, obviously? Hold on to that answer. Today one idea settles it in a single line: conservation of mechanical energy. It comes up often in JEE, so let's begin.

0:32When does mechanical energy stay constant?

So, mechanical energy. Capital E equals capital K plus capital U, okay? K is the kinetic energy, the K E, and U is the potential energy. So E is their total. Now tell me. Does this total stay the same in every motion? No, not in every motion! Only when the forces doing work are conservative, like gravity. Then whatever potential energy is lost becomes kinetic energy, and the total stays fixed. Right? Now, a track also pushes on the body: the normal reaction. On a fixed smooth surface, with no friction, this push is perpendicular to the velocity, the upside down T on screen. So it does no work. Clear? Same with a pendulum: the string's tension is perpendicular to the bob's velocity, so no work either. So on a fixed smooth track, or for a pendulum, E stays constant. Got it?

1:26Smooth track: only the height matters

Now let's use it. Take a body of mass small m on a smooth track, starting with speed small u, and let it slide down a height h. With zero potential energy at the bottom: at the top, E is half m u squared plus m g h, g being the acceleration due to gravity; at the bottom, half m v squared, v the final speed. E is constant, so they are equal. Okay? Now look: every term has m, so m cancels! v equals the square root of, u squared plus two g h; the tick shaped sign is the square root. Really simple, right? Now think. Two smooth slides from the same height, one short and steep, one long and curvy, and the same starting speed. At the bottom, which body is faster? Neither! They arrive at the same speed. If you said that, yes, exactly! The formula has no angle, no track length, and no mass. Any shape, any mass, the same v: only the height fallen matters, as long as the track is smooth. Okay?

2:29Pendulum: speed at the lowest point

Next, the pendulum: a bob on a string of length small l, pulled aside to angle theta naught with the vertical and released from rest. Theta is the Greek letter; the small zero below it is read naught. With me? The tension does no work, so E is conserved, and we only need the height fallen. So, how far does the bob drop to the lowest point? At the start, the bob is l cos theta naught below the point of suspension; at the lowest point, the full length l below. So it falls l minus l cos theta naught, that is l times, one minus cos theta naught. Okay? Put this h into root two g h: the speed at the lowest point, v with the small word lowest below it, is the square root of two g l times, one minus cos theta naught. It starts from rest, so no u squared. Got it?

3:20Chains and ropes: use each part's centre of mass

Now, chains and ropes. They are long bodies, so take each part's potential energy from the height of its own centre of mass. Take a chain of mass capital M and length capital L on a table. In the figure, a length small l hangs over the edge, and L minus l stays on the table. Clear? Now, the hanging part. Where is its centre of mass? Halfway down, l over two below the table top. Yes, exactly, the orange dot in the figure; C M there stands for centre of mass! And mass goes with length, so the hanging part has mass M l over L, okay? So, the work to pull the hanging part back onto the table is the potential energy it gains: M l over L, times g, times l over two. That is M g l squared over two L, right? And a topper's trick. If the table and its edge are smooth and the chain is released from rest, it leaves the table at v equals the square root of, g times, L squared minus l squared, over L, the full length. We will use it as a check soon. With me?

4:25Example 1: stone thrown from a cliff in three directions

Back to our puzzle: Example one. Here, a stone is thrown from the top of a twenty five metre cliff at ten metres per second; m s to the minus one means metres per second. Find its speed on reaching the ground if it is thrown horizontally, at thirty degrees above the horizontal, the little circle being degrees, or straight down. Take g as ten. Pause the video, and try it. Done? Okay, let's check. In the air, only gravity does work, so E is conserved, and v squared equals u squared plus two g h. Look, no angle anywhere! Right? So now tell me: which stone hits the ground fastest? None of them! They all land at the same speed. Ten squared is one hundred; two times ten times twenty five is five hundred. So v squared is six hundred, whatever the direction. Okay? So v is root six hundred, about twenty four point five metres per second; the wavy equals sign means approximately. The same in all three cases. Puzzle solved! The direction does not matter. Clear?

5:33Example 2: speed of a chain sliding off a smooth table

Now a harder one, Example two. Here is a two kilogram uniform chain, one metre long; k g means kilograms. It lies on a smooth table, point four metres hanging over the edge, and is released. Find its speed as the last link leaves the table, with g as ten. Pause and try. Hint: each part's centre of mass. Done? Okay, let's check. The table is smooth and only gravity does work, so E is conserved. Take the table top as zero potential energy; anything below it is negative. Okay? At the start, point four of the one metre hangs, so the hanging mass is point four of two kilograms: point eight k g. Its centre of mass is halfway down: point two metres. Right? So U i, U with a small i for initial, is minus point eight times ten times point two: minus one point six joules, capital J for joules. The part on the table sits at zero. Clear? At the end, the whole chain hangs, with its centre of mass half a metre down. So U f, with a small f for final, is minus two times ten times point five: minus ten joules. So the kinetic energy gained is ten joules, right? No, no, careful! That is the trap. The chain did not start at zero; it already had minus one point six joules. Kinetic energy gained equals potential energy lost: U i minus U f, minus one point six minus minus ten, eight point four joules. And half times two times v squared is just v squared. Okay? So v is root eight point four, about two point nine metres per second. Now the topper's check: L is one and l is point four, so g times, L squared minus l squared, over L is ten times, one minus point one six, point one six being point four squared. Eight point four again. Same answer!

7:31Quick revision: K + U constant, track, pendulum, chain

So, quick revision. When only conservative forces do work, E equals K plus U stays constant. Smooth track: v is the root of u squared plus two g h, any shape, any mass. Pendulum: the root of two g l times, one minus cos theta naught. Chains and ropes: each part's potential energy from its own centre of mass. Now try the practice problems on this page. You can do them. All the best!