Part C: Advanced-Level Problems

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a uniform rod lying on a smooth table. You give one end a sharp sideways hit. The end you hit moves forwards, of course. But the other end? Forwards too, right? Hmm, are you sure? Hold that thought; we settle it in Example one. Today is Part C of the unit problem bank: longer problems at JEE Advanced level, each joining more than one idea. So, let's begin.

0:32Part C: two or three ideas in every problem

So, here is Part C: four longer problems at JEE Advanced level, each needing two or three ideas of the unit at once. Okay? The hinge force on a falling rod, a free rod hit by an impulse, a sphere rolling on a pulled plank, and a bead moving on a free ring. Today we solve the impulse and the plank problems, with a topper's check from the hinge problem. Right?

0:57Idea 1: a free rod gets an impulse

Now, the first idea. Say a free rigid body at rest gets a sharp blow, an impulse capital J, okay? First, the linear impulse moves its centre of mass: J equals m times v c m, where m is the mass, and v c m, with the small c m below, is the velocity of the centre of mass. Clear? Second, the body also starts to spin. About which point is it easiest to take the angular impulse? About the centre of mass, yes, exactly! Angular impulse equals the change in angular momentum: J times d equals I c m times omega. Here d is the impulse's perpendicular distance from the centre, I c m the moment of inertia about the centre of mass, and omega, the Greek letter, the angular speed. Got it? Now combine the two. For a rod, a point at distance r from the centre moves at v c m, plus or minus omega r. Plus where the spin carries the point forwards, minus where it carries it back. Okay?

2:03Idea 2: rolling on a moving surface

Now the second idea. Rolling without slipping means the contact point moves together with the surface below it, with no sliding. Clear? On fixed ground the contact point is at rest, so a equals alpha R, okay? Here a is the acceleration of the centre, alpha, the Greek letter, the angular acceleration, and capital R the radius. Right? Now put a sphere on a plank that is itself moving. Is a equals alpha R still true? No, no, careful! That is the trap. The contact point must now move with the plank. So a p minus a s equals alpha R; the small p below means the plank, and the small s, the sphere. The spin makes up only the difference. Got it?

2:47Topper's trick: check at special positions

Now a topper's trick, from the hinge problem. Here a uniform rod, hinged at one end, turns about that fixed axis after release from the horizontal. When it has turned through an angle theta, the Greek letter, from the horizontal, the vertical hinge force, capital V, is one quarter m g times one plus nine sine squared theta. The horizontal part, capital H, is nine quarters m g sine theta cosine theta. Long answers! So check them at special positions. Okay? So, put theta equal to zero, the moment of release. What does V become? Sine zero is zero, so V is just one quarter m g. Correct! And it makes sense: at release the centre already accelerates down at three quarters g, so the hinge holds up only a quarter of the weight. Very good! Now the vertical: theta equals ninety degrees; the small circle means degrees. Sine is one and cosine is zero, so V is one quarter m g times ten, five over two m g, and H is zero. Exactly what Part A found for this rod at the vertical! Two checks pass, so trust the answer. Clear?

4:03Example 1: impulse at one end of a free rod

Now, our puzzle from the start, as a Part C problem. Read it with me: a uniform rod of mass m and length capital L lies on a smooth horizontal table. An impulse capital J is applied at one end, perpendicular to the rod. Find the velocity of the other end just afterwards. Pause the video, and try it. Done? Okay. Many students say: the whole rod gets the impulse, so the far end moves forwards at J over m. Right? No, no, careful! J over m is only the velocity of the centre of mass: v c m equals J over m. The rod also spins, and we must add that. Okay? Now the spin, about the centre. The impulse acts at distance L over two, and for a rod about its centre, I c m is one over twelve m L squared. So J times L over two equals one over twelve m L squared omega. The double arrow means this gives: omega equals six J over m L, right? Now the far end. It is on the other side, so the spin carries it backwards: minus omega times L over two. That is J over m, minus three J over m. Minus two J over m! With me? The minus sign means backwards: opposite to J, with speed two J over m. Puzzle solved: hit one end, and the other end jumps back! And the point L over six from the centre, towards the far end, is at rest for that moment. Very good!

5:41Example 2: sphere rolling on a pulled plank

Now a harder one. Read it with me: a solid sphere of mass m and radius capital R rests on a plank of mass capital M, lying on a smooth floor. The sphere rolls on the plank without slipping, and the plank is pulled with a horizontal force capital F, okay? Find the accelerations of the plank and of the sphere. Pause the video, and try it. Done? Okay, let's check. The sphere touches only the plank, so its only horizontal force is friction. Which way does it push the sphere, forwards or backwards? Forwards, yes! The plank drags the sphere along. So friction, small f, acts forwards on the sphere and backwards on the plank. Sphere: f equals m a s. And torque about its centre: f R equals two fifths m R squared alpha, where two fifths m R squared is I for a solid sphere about its centre. Okay? Now the rolling condition, and our trap from before: the plank moves, so a s equals alpha R is wrong here. The contact point moves with the plank: a p minus a s equals alpha R, right? From the torque equation, alpha R equals five f over two m, and since f over m is a s, that is five over two a s. So a p equals a s plus five over two a s: seven over two a s. Got it? Now the plank: capital F forwards, small f backwards, so capital F minus f equals capital M a p. Put f equals m a s and a p equals seven over two a s: capital F equals m a s, plus seven over two capital M a s. With me? So a s equals two capital F, over two m plus seven capital M, and a p equals seven capital F, over the same denominator. And see, a p is seven over two times a s, just as our rolling line said. Perfect!

7:46Quick revision: impulse, moving surface, special positions

So, quick revision. When a free body is hit by an impulse capital J, J equals m v c m, plus angular impulse about the centre of mass, okay? When a body rolls on a moving surface, its contact point moves with that surface. And for a long answer, check it at special positions. Now try the hinge and the bead problems on this page, then the practice problems. You can do this. All the best!