Part B: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small race before we start. Take a ring and a disc with the same mass and the same radius, and let them roll down the same slope together. Which one reaches the bottom first? And which one has more kinetic energy there? Same mass, same radius, same slope, so it's a tie, right? Hmm, are you sure? We will settle it in Example two. Today: concept checks on the whole unit of rotational motion. These short statement questions come up often in JEE, and one word can flip the answer. So, let's begin.

0:37Two statements: judge each one separately

So, first, two statements, marked with the Roman numerals one and two. One: the centre of mass of a body always lies inside it. Two: the centre of mass of a system of particles moves only under external forces. The method on screen: check each one separately. Okay? Look at the figure: four standard bodies, each red dot a centre of mass, and the green arrows give its distance from the centre, capital O, or from the base. The first is a semicircular ring, shaped like a horseshoe. So, does the centre of mass always lie inside the body? No, not always! Here the dot is two R over pi above O, where R is the radius and pi is the Greek letter. No material there, just empty space. And for a full ring, the centre of mass is at the middle, again empty. So statement one is false. Right? Now statement two. The particles push and pull on each other: internal forces. Can internal forces change how the centre of mass moves? No, they cannot! They come in equal and opposite pairs, by Newton's third law, so they cancel. What is left: F ext equals capital M times a c m. The arrows on top mean vectors, ext means external, capital M is the total mass, and a c m is the acceleration of the centre of mass. So only external forces can change the motion of the centre of mass; with no external force, it just keeps moving at constant velocity. In that sense, statement two is true. Clear?

2:19Angular momentum: conserved about which point?

Next, angular momentum, capital L, okay? Three cases on screen: a planet round the Sun, a bullet stuck in a hinged rod, and a rolling ball. The rule: if the external torque about a point is zero, then L about that point stays constant; the double arrow means so. So always ask: about which point? With me? First case. The Sun's gravity pulls on the planet. Does it have any torque about the Sun? None, exactly! Gravity points straight at the Sun, a central force, so its line passes through the Sun and its torque about the Sun is zero. So yes, L about the Sun is conserved. Got it? Second case: a bullet gets stuck in the rod. During the hit, the hinge pushes hard on the rod, an external force. But it acts at the hinge itself, so about the hinge it has no torque. So yes, but only about the hinge. Okay? Third case: a ball rolling down an incline. Gravity and the normal force both pass through its centre, so no torque from them. So L about the centre is conserved, right? No, no, careful! You forgot friction. It acts at the contact point, one radius from the centre, so it has a torque about the centre. That torque makes omega, the Greek letter for the angular speed, increase as the ball rolls down. So L changes, and the answer is no. Clear?

3:46Rolling: does static friction do work?

Now, the question on screen: does static friction do work on a body rolling without slipping on a fixed surface? Look at the figure: rolling is translation plus rotation. In translation, every point moves forward at v, the speed of the centre. In rotation, the rim moves at omega R, with R the radius: forward at the top, backward at the bottom. Rolling without slipping means v equals omega R, okay? So the bottom point has v forward, and omega R backward. What is its velocity? Zero, yes! v minus omega R is zero. That is the contact point, capital P, the red dot, at rest at that instant. And the other arrows: the top moves at two v, and the point level with the centre at root two v. Right? Static friction acts exactly at P, the contact point. So, does it do any work? No work at all! Work needs the point where the force acts to move, and P does not move. Clear? But friction is not useless. Remember the rolling ball? Its torque made omega increase. So friction shifts energy between translation and rotation. Got it? It takes from one and gives to the other, so the total mechanical energy stays unchanged. We need this in Example two. With me?

5:10Example 1: does moment of inertia depend on omega

Alright, Example one, an easy one. Does the moment of inertia of a body depend on its angular velocity? On the axis? On the distribution of mass? Pause the video, and try it first. Done? Okay, let's check. Capital I, the moment of inertia, equals the sum of m i r i squared. The Greek sigma means sum, the small i written low labels each particle, m i is its mass, and r i its perpendicular distance from the axis. Now, spin the body faster. Does I change? No, it does not! Omega is nowhere in the formula. Many students fall for this one. Okay? But change the axis, and every r i changes. Move the mass closer to the axis or farther, and again the r i change. So yes, and yes. That is why we always state the axis. Right? So the answer: no, yes, yes. One body, one mass, but many moments of inertia. Very good!

6:14Example 2: ring vs disc down an incline, which wins

Now our race, Example two. Two bodies of the same mass and radius, one a ring and one a disc, roll down the same incline. Which has the greater kinetic energy at the bottom? Which reaches the bottom first? Pause the video, and try it. Done? Okay, let's check. Kinetic energy first. Which one has more at the bottom? Here is the key. Remember: in rolling without slipping, static friction does no work. So mechanical energy is conserved. Clear? Both drop the same height, h, so both lose the same potential energy, capital M g h, where g is the acceleration due to gravity. The double arrow means so: the same kinetic energy, capital K, at the bottom. So, do they also arrive together? No, no, careful! The time depends on the acceleration. For rolling without slipping down an incline, the acceleration, a, equals g sine theta, over one plus k squared over R squared. Here theta, the Greek letter, is the angle of the incline, k is the radius of gyration, and R is the radius. Okay? For a ring, all the mass is on the rim, so k equals R, and k squared over R squared is one. So a is one half g sine theta. With me? For a disc, k squared over R squared is one half. So a equals g sine theta over one and a half: two thirds g sine theta. Two thirds is more than one half, so the disc accelerates faster. Right? So: equal kinetic energies, and the disc arrives first. Not a tie after all! Same energy at the bottom, but the disc wins the race. That's it!

8:04Quick revision: four rules on CM, L, friction and I

So, quick revision. One: judge each statement separately; the centre of mass can lie outside the body. Two: angular momentum, L, is conserved about a point when the external torque about it is zero. Three: in rolling without slipping, static friction does no work. Four: capital I depends on the axis and the mass distribution, not on omega. Now go and try the practice problems on this page. You can do them. All the best!