Friction in Rolling; Slipping That Becomes Rolling

Transcript

0:00Introduction

Hello everyone! Throw a ball along a rough floor so that it slides, with no spin. In a moment, it is spinning, and then it is rolling! Who made it spin? Friction, of course. But wait, friction slows things down, right? So how does it make the ball spin faster? And when the ball finally rolls, how fast is it moving? By the end, you will answer it in two lines. Friction in rolling comes up often in JEE, so let's begin.

0:31Rolling with friction: three equations

So, first idea. Say a rigid body rolls without slipping on rough ground. Now tell me, which way does static friction act on it? Backwards, against the motion? That is what most students say. Oh no, careful! Static friction, small f, has no fixed direction. So, assume a direction, and solve. If f comes out negative, it simply acts the other way. Okay? Equation one: sigma F equals M times a c m, where capital M is the mass. The big Greek letter sigma means sum, so this is the sum of the forces, and a c m, with the little c m below, is the acceleration of the centre of mass. Right? Equation two, for the turning: sigma tau c m equals I c m times alpha. Tau is torque, and I c m is the moment of inertia, both about the axis through the centre of mass, and alpha is the angular acceleration. With me? But look, two equations and three unknowns: a c m, alpha and f. The body rolls without slipping, so what connects a c m and alpha? Yes, exactly! The rolling condition: a c m equals alpha times capital R, the radius. True only for rolling without slipping, clear? One last check. Pure rolling is possible only if the required friction, its size shown between the two straight bars, is less than or equal to mu N, where mu is the coefficient of friction and capital N is the normal force. Need more than that, and the body slips. Got it?

2:10Slipping that becomes rolling

Second idea. This figure is from the kinematics of rolling. Left: translation, every point moves forward at v. Middle: rotation with angular speed omega, the Greek letter by the curved arrow, so the bottom point moves backwards at omega R, okay? Add them to get rolling, on the right: the top moves at two v, the side at root two v. When v equals omega R, the bottom point, the red dot P, is at rest. Right? But suppose v is not equal to omega R, the equals sign with a slash through it. So what does the bottom point do now? It slides on the ground! Correct. The body is slipping, so the friction is kinetic, with its full value, mu times capital N, okay? This kinetic friction keeps changing both v and omega, until v equals omega R, and from then on, the body rolls. Clear? And on level ground, with no other force pushing it along, a rolling body needs no friction at all. So once it rolls, the friction drops to zero, and it simply keeps rolling. With me?

3:18Shortcut: angular momentum about the contact point

Now, the final speed. Friction has a torque about the centre, so angular momentum about the centre keeps changing. Is there a point about which friction has no torque at all? The contact point! Yes, exactly. Friction passes right through it, so its torque about that point is zero. Okay? No torque, no change. So on level ground, capital L, the angular momentum about the contact point, is conserved during the slipping. This is the topper's shortcut. Got it? Before: M v zero R from the forward motion, plus or minus I c m omega zero from the spin. The little zero below means the starting value. After, the body rolls: M R squared plus I c m, in brackets, times v over R, right? Now the plus or minus sign. Plus when the starting spin is in the forward-rolling sense, and minus for backspin.

4:16Example 1: force at the centre of a rolling cylinder

Clear? Alright, example one. Here, a horizontal force, capital F, acts at the centre of a solid cylinder of mass capital M, on rough ground. Find its acceleration, and the friction force, if it rolls. Pause the video, and try it first. Done? Okay, let's check. Capital F acts right at the centre, so its torque about the centre is zero. Then what can make the cylinder spin? Only friction! So friction acts backwards at the bottom, to turn the cylinder forwards. Along the ground: capital F minus small f equals M a, where a is the acceleration of the centre. With me? Now torques about the centre, where only friction has a torque: f times R, with R the radius, okay? For a solid cylinder, I c m is half M R squared, and rolling gives alpha equal to a over R, so f R equals half M a R, right? The double arrow means so: small f equals half M a. Clear? Put this into the first equation: capital F equals three halves M a. So a equals two F over three M, and small f equals F over three, backwards. With me? Now our check. The force is horizontal, so N equals M g, where g is the acceleration due to gravity. Pure rolling needs F over three to be less than or equal to mu M g, so capital F at most three mu M g. Push harder, and the cylinder slips. Got it?

5:56Example 2: ball thrown with no spin starts rolling

Now the harder one: our opening puzzle! Here, a solid ball is thrown along a rough floor, with mu equal to zero point two, at seven metres per second, and no spin. Find its speed when it starts rolling, the time taken, the distance slid, and the fraction of kinetic energy lost. Take g as ten. Hint: angular momentum about the contact point. Pause the video and try it. Done? Let's check. Angular momentum about the contact point. Before: no spin, so only M v zero R, okay? After: it rolls, and for a solid sphere, I c m is two fifths M R squared. The dot means times. So M v zero R equals M v R, plus two fifths M R squared times v over R, right? Cancel M R from every term. v equals five sevenths of v zero: five sevenths of seven, five metres per second. Clear? Now, how long does the sliding last? Which force slows the ball down while it slides? Kinetic friction, mu M g! Very good. Divide by the mass: the deceleration is mu g, zero point two times ten, two metres per second squared. Speed falls from seven to five, so t is seven minus five, over two: one second. Got it? Distance slid: seven times one, minus half, times two, times one squared. Six metres. With me? Now energy. K E is kinetic energy. At the end, it is just half M v squared, right? No, no, wait! The ball is also spinning now. For a rolling solid sphere, it is half M v squared, times seven fifths. Put v as five sevenths of v zero, and it becomes five sevenths of the starting energy, half M v zero squared. Okay? So the fraction lost is one minus five sevenths: two sevenths. Puzzle solved: friction slowed the ball from seven to five, and spun it up at the same time, until it rolled. That's it!

8:07Quick revision: friction equations, slipping to rolling

So, quick revision. Rolling with friction: sum of forces equals M a, sum of torques equals I alpha, and a equals alpha R, then check that friction is at most mu N, okay? When a body slips, kinetic friction acts until v equals omega R, and then it rolls. And for the final speed, conserve angular momentum about the contact point. Now go and try the practice problems on this page. All the best!