Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Picture a uniform rod spinning freely in a horizontal plane, about a vertical axis through its centre. No motor, no friction, nobody touches it. Now two small beads slide out from its centre to its ends, and it slows down! So who applied the brakes? Strange, right? We will settle it at the end. This is set one of mixed problems on rotational motion, in the JEE Main pattern: two quick ideas, then two full problems. So, let's begin.

0:37Idea 1: a disc with a hole

Idea number one: moment of inertia. Here, a uniform disc of radius two R has a concentric hole of radius R, with the same centre. Its mass is capital M, and its moment of inertia about its axis is x over two, times M R squared. Find x, okay? There is a hole, so subtract it, as in the badge: half M, times four R squared minus R squared. Three halves M R squared, so x is three, right? No, no, careful! Capital M is the mass of the plate that is left, not of the full disc, so we cannot subtract like that. Use the standard result for an annular disc, a ring-shaped plate with radii R one and R two; the small one and two below are labels. About its axis, I, the moment of inertia, equals half M, times R one squared plus R two squared. See the plus sign! Clear? Here R one is R, and R two is two R, okay? So I equals half M, times R squared plus four R squared: five halves M R squared. Match it with x over two, M R squared. The double arrow means so: x over two is five over two, so x equals five. Five, not three! Got it?

1:49Idea 2: a block on a heavy pulley

Alright, idea number two: torque with a pulley. Here, a five kilogram block hangs from a string wound on a ring-shaped pulley, also five kilograms, turning on a fixed axle. Find the acceleration and the tension. With me? The block hangs from the string. So the tension, capital T, is just its weight, m g: fifty newtons, where capital N is the newton, right? Oh no, careful! If T were m g, the block would feel no net force, and could not speed up. The string does not slip on the pulley, so the block's acceleration, a, equals alpha, the pulley's angular acceleration, times capital R, its radius. Okay? Newton's second law for the block, down positive, with g as ten metres per second squared, written m s to the power minus two. Small m is the block's mass: m g minus T equals m a. Right? The pulley is a rigid body on a fixed axle, so torque equals I alpha, and T times R is the torque. For a ring, all the mass is on the rim, so I equals capital M R squared. Put alpha as a over R, and the double arrow gives T equals capital M a. Clear? Add the two equations, and T cancels: m g equals, m plus capital M, times a. So a equals m g over, m plus capital M, that is fifty over ten: five metres per second squared. Got it? And T equals m times, g minus a: five times, ten minus five, twenty five newtons. Only half the weight, not fifty! And capital M a, five times five, agrees. With me?

3:24Example 1: kinetic energy of a rolling solid cylinder

Now, example one, an easy one. Here, a two kilogram solid cylinder rolls at four metres per second; m s to the power minus one means metres per second. Find its kinetic energy. Pause the video, and try it first. Done? Okay, let's check. Half M v squared: half, times two, times four squared, sixteen joules, where capital J is the joule, right? No, no, wait! Look at the figure. On the left, translation: every point moves forward at v. In the middle, rotation: the body spins at angular speed omega, by the curved arrow, and the rim moves at omega R, okay? Add them, and on the right is rolling, with v equal to omega R: the red contact point P is at rest, the top moves at two v, the side at root two v. So rolling has two kinetic energies, and sixteen joules is only the first. Capital K, the total, is half M v squared for the translation, plus half I c m omega squared for the rotation. I c m, with c m below, is the moment of inertia about the axis through the centre of mass. Right? Rolling without slipping gives omega equals v over R, with R the radius, and we write I c m as M k squared, where k is the radius of gyration. Clear? Put both in: K equals half M v squared, times one plus k squared over R squared. The one is the translation; k squared over R squared is the spin's share. With me? For a solid cylinder, I c m is half M R squared, so k squared over R squared is one half. Got it? So K equals half, times two, times sixteen, times one plus a half. Sixteen times one point five: twenty four joules. The trap answer, sixteen, was only the translation. Perfect!

5:15Example 2: beads slide out on a spinning rod, new speed

Now the harder one: our puzzle from the start. Here, a uniform rod of mass capital M and length capital L spins freely at omega, in a horizontal plane, about a vertical axis through its centre. Two beads of mass small m each start at the centre and slide out to the two ends. Find the new angular speed. Pause the video and try it. Hint: is there any torque? Done? Okay, let's check. The rod spins freely, and the beads push only on the rod, from inside. So no external torque. Then what stays constant? Angular momentum, exactly! Internal forces cannot change it. So I one omega equals I two omega prime: I one and I two are the moments of inertia before and after, and omega prime, with the small dash, is the new angular speed. Okay? Now, before. The beads sit right at the centre, on the axis. So how much do they add to I one? Nothing, correct! Their distance from the axis is zero. So I one is just the rod's own: one twelfth M L squared, about its centre. Clear? Now, after. The beads are at the two ends. So each bead is a distance L from the axis, right? Oh no, careful! The axis is at the centre, so each end is only L over two away. Two beads: two m, times L over two, squared. So I two is one twelfth M L squared, plus half m L squared. With me? Now conserve it: one twelfth M L squared times omega, equals the bracket times omega prime. Got it? Multiply both sides by twelve over L squared: omega prime equals M omega, over M plus six m. So, our puzzle: nobody braked the rod! The beads moved mass away from the axis, so I went up. With no external torque, angular momentum stays the same, so omega had to come down. That's it!

7:13Quick revision: annular disc, pulley, rolling KE, L = const

So, quick revision. Annular disc: I equals half M, times R one squared plus R two squared, with a plus sign. Block on a pulley: a equals m g over, m plus I over R squared; for a ring, I over R squared is capital M, okay? Rolling: K equals half M v squared, times one plus k squared over R squared, never just half M v squared. No external torque: I one omega one equals I two omega two. Now go and try the practice problems on this page. All the best!