Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Hold a solid cylinder with its axis horizontal, and spin it fast about that axis. Now lower it gently onto rough, level ground. First it skids, then it rolls forward. So here is my question: by the time it rolls, how much of its kinetic energy is lost? Just a little? Half? Keep your guess ready. This is set three of mixed problems on rotational motion, JEE Main pattern: two ideas, three problems from the course, and the last one is this puzzle. So, let's begin.

0:39Idea 1: a disc turning about a rim point

So, idea number one, a problem from this set. Here, a uniform disc of radius capital R can turn freely in a vertical plane, about a horizontal axis through a point on its rim. It is released with its centre level with the axis. Find its angular speed when the centre is directly below the axis. First, the moment of inertia, I, about this axis. For a disc, it is half m R squared, with small m the mass, right? No, no, careful! Half m R squared is about the axis through the centre. Our axis is on the rim, parallel to that one, at a distance capital R from the centre. So, by the parallel axis theorem, add m R squared. Half plus one: three halves m R squared. Remember this number; it comes back today. Got it? Now omega, the Greek letter, the angular speed. The axis is fixed, with no friction. So which law should we use? Conservation of energy, yes! The centre falls through a height capital R, so the potential energy lost is m g R, with g the acceleration due to gravity. All of it becomes rotational kinetic energy: half, times three halves m R squared, times omega squared. The dot in the middle means times. Clear? Cancel m, and one power of R, from both sides. Omega squared equals four g over three capital R, and the double arrow means so. Omega is the square root of four g over three R, the tick shaped sign being the square root. With me?

2:13Idea 2: rolling turns about the contact point

Now, idea number two, for rolling. Look at the figure. Left: translation, every point moves forward at v. Middle: rotation about the centre, with angular speed omega, shown by the curved arrow, so the bottom point moves backwards at omega R, as shown. Add them: on the right, rolling, where the top moves at two v, and the point on the side at root two v, along the slanting arrow. When v equals omega R, the body rolls without slipping, and the contact point, the red dot P, is at rest. Okay? Now think. At this instant, the point P is not moving at all. So, about which point is the whole body turning? About P itself, exactly! The contact point is the instantaneous axis. So we need I P, the moment of inertia about it, with the little P written below. Parallel axes again: I P equals ICM, the little c m below meaning the axis through the centre of mass, plus capital M R squared, with capital M the mass. For a disc: three halves capital M R squared. The same number as idea one, because P is also a point on the rim! Right? So, about P, rolling looks like pure rotation. L P, the angular momentum about P, equals I P omega, and the kinetic energy, capital K, equals half I P omega squared. Clear? Now, if v is not equal to omega R, the body slips, and kinetic friction acts at P, along the ground, until it rolls. So, what is the torque of this friction about the point P? Zero, correct! The friction passes right through the point P, so no torque. And gravity and the normal force act along the same vertical line through P, so they give no torque either. So, on level ground, the angular momentum about the contact point is conserved while it slips. We need this in the last problem. Got it?

4:14Example 1: angular momentum of a rolling disc

Alright, example one. Here, a disc of mass capital M and radius capital R rolls with speed v. Find its angular momentum about, part a, the contact point, and part b, any fixed point on the ground. Pause the video, and try it first. Done? Okay, let's check. Part a is idea two, directly. L P equals I P omega. For a disc, I P is three halves M R squared, and for rolling, omega is v over R, so multiply: three halves M R v. Right? Now part b, a point fixed on the ground. The disc moves forward and spins at the same time. So its angular momentum has two parts. Which two? Orbital and spin, exactly! The orbital part comes from the centre moving, as if the whole mass sat there. The spin part comes from turning about the centre. With me? Orbital first. The centre moves at speed v, along a line at height capital R above the ground. So about our fixed point, the orbital part is M v times R, okay? Wherever the point is on the ground, that height stays the same. Now spin: ICM times omega. Half M R squared, times v over R: half M R v. Clear? Both parts turn the same way, so they add. M v R plus half M R v is three halves M R v. The same as part a! So for a rolling disc, it is three halves M R v about any fixed point on the ground. Got it?

5:53Example 2: spinning cylinder on rough ground, energy lost

Now the harder one: our puzzle from the start. Here, a solid cylinder spinning at omega zero, the little zero below meaning the starting value, is placed on rough level ground. What fraction of its kinetic energy is lost by the time it rolls? Hint: idea two. Pause the video and try it. Done? Okay, let's check. First we need the final speed, v. Many students set the starting spin energy equal to the final rolling energy. Is that allowed? No, no, careful! While it skids, kinetic friction acts, and energy is lost. That loss is our answer! So use idea two: angular momentum about the contact point. Before, the centre is at rest, so there is only spin: half M R squared, times omega zero. After, it rolls. Its ICM is the same as a disc's, so example one gives three halves M R v. With me? Cancel M R from both sides. v equals omega zero R over three. Right? Now the energies. K one, with the little one below, is the starting kinetic energy, all spin: half, times half M R squared, times omega zero squared. That is one quarter M R squared omega zero squared. Clear? Now K two, the kinetic energy at the end. It is just half M v squared, right? Oh no, careful! The cylinder is still spinning as it rolls. So it is half M v squared, times three halves, to include the spin. Put v as one third omega zero R: three quarters M, times omega zero squared R squared over nine. That is one twelfth M R squared omega zero squared. Okay? So K two over K one, the slash meaning over, is one twelfth over one quarter: one third. Only one third is left, so two thirds is lost! More than half, all while it skids. Puzzle solved!

7:51Quick revision: disc about rim, rolling L, skid to roll

So, quick revision. For a disc about a rim point, with the axis perpendicular to the disc, I equals three halves M R squared. For a rolling disc, the angular momentum about any fixed point on the ground is three halves M R v. And from skidding to rolling, conserve angular momentum about the contact point, not energy. Now go and try the five practice problems on this page. All the best!