Collisions Involving Rotation

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Picture a rod hanging from a hinge, and a bullet that flies in and gets stuck in its lower end. Your hand already wants to write: linear momentum before equals linear momentum after, right? But here, that gives the wrong answer! Something in this picture is breaking the rule. Can you guess what? By the end of this video, you will know what it is, and which law to use instead. Collisions with rotation come up often in JEE, so let's begin.

0:34Body on a hinge

So, look at the figure. The rod has mass capital M and length capital L, and it hangs from a hinge, the red dot at the top. Careful: here capital L is the length, not angular momentum, okay? The green bullet, of mass small m, comes in with speed v, the green arrow. After the hit, the rod swings with angular speed omega, the Greek letter beside the curved blue arrow. And the equation on the right? We will build it ourselves in a moment. Now tell me: during the impact, is linear momentum conserved? No, it is not! The hinge holds the top end of the rod in place. To do that during such a short impact, the hinge has to exert a very big force on the rod. That is an outside force on the rod and the bullet together, so their linear momentum is not conserved. That is what breaks the rule in our puzzle! Got it? So, linear momentum fails. But wait. Torque is r cross F, remember? So tell me: what is the torque of the hinge force, taken about the hinge itself? Zero! Yes, exactly. The hinge force acts right at the hinge, so r, measured from the hinge, is zero. So tau, the Greek letter we use for torque, is zero about the hinge. Right? And about the hinge, no outside torque acts during the impact, so angular momentum about the hinge does not change. So, for a body on a hinge: conserve angular momentum about the hinge. Now look at the equation on the right of the figure. It writes this for our rod: m v L equals, in brackets, one third capital M L squared plus m L squared, times omega. We will build every term of it in example one. With me?

2:21Free body: no hinge

Alright, now take the hinge away. Think of a rod lying free on a smooth table, so no friction, and no hinge. Something hits it. Now which quantities are conserved? Linear momentum, yes! No hinge means no big outside force during the impact. So linear momentum is conserved. Clear? And angular momentum too. You can take it about the centre of mass, or about any fixed point. So a free body gives you two equations, not one. We will use the centre of the rod in example two. Okay?

2:55Energy: only for an elastic collision

Now, the third law to think about: energy. Kinetic energy is conserved only if the question says the collision is elastic. Got it? So, back to our bullet and rod. Many students write: half m v squared equals half I omega squared, where I is the moment of inertia. Looks fine, right? No, no, careful! This is the trap. The bullet embeds in the rod. That collision is not elastic, so part of the kinetic energy is lost. So, no energy equation here. Use angular momentum about the hinge instead. With me?

3:31Example 1: bullet embeds in a hinged rod

Now, example one. Here, a bullet of mass m, moving horizontally with speed v, embeds in the lower end of a uniform rod, of mass capital M and length capital L, which hangs from a hinge at its top. Find the angular speed just after the impact. This is the rod from our figure. Pause the video, and try it first. Done? Okay, let's check. Hinge, so angular momentum about the hinge. Before the impact, only the bullet moves. Its angular momentum is momentum times perpendicular distance, remember? The bullet moves horizontally at the lower end, a distance L below the hinge, so it is m v times L, okay? Just after the impact, the rod and the bullet turn together, with angular speed omega. What is the moment of inertia now? Two parts! First the rod. It turns about a fixed axis through its end, not its centre, so one third capital M L squared. Plus the bullet, a point mass at distance L from the hinge, so m L squared. Students often forget this second term, but the bullet turns too! Clear? So, angular momentum before equals angular momentum after. m v L equals, in brackets, one third capital M L squared plus m L squared, times omega. The same equation as in the figure. Right? Now solve. Cancel one L from both sides, and multiply the top and the bottom by three. Omega equals three m v, over, capital M plus three m in brackets, times L, that's it! Notice: no linear momentum, no energy. One law, one line. Got it?

5:16Example 2: ball hits a free rod elastically, find M/m

Now a harder one, a JEE pattern question. Here, a ball of mass m, moving with speed v on a smooth table, hits one end of a free uniform rod of mass capital M, at right angles. The collision is elastic, and the ball stops dead. Find capital M over m. Hint: a free rod, and an elastic collision. Pause the video and try it. Done? Okay, first things first. Which conservation laws can we use here? All three! Yes, exactly. Free rod, so linear momentum. Elastic, so energy. And angular momentum, about the centre. Start with linear momentum. The ball stops dead, so all of m v goes to the rod: m v equals capital M times v c m. Here v c m, with the small subscript c m, is the speed of the rod's centre of mass. The double arrow means so: v c m equals m v over capital M, with me? Now angular momentum, about the centre of the rod. Let the rod's length be L, okay? In example one we used one third capital M L squared. Same here, right? No, no, wait! One third was about the end, where the hinge was. Here there is no hinge, and we take the centre, so it is one twelfth capital M L squared. Before: the ball's line of motion is half the length from the centre, so m v times L over two. After: the ball has stopped, so only the rod turns. That gives omega equals six m v over capital M L, clear? Now energy, because the collision is elastic. Before: half m v squared. After: the ball has nothing, and the rod has two kinds of kinetic energy. Half capital M v c m squared for moving, plus half I omega squared for turning, with I equal to one twelfth capital M L squared. The small dot just means times. Okay? Multiply by two: the left side is m v squared. Put in v c m and omega. The first term becomes m squared v squared over capital M, with me? The second: one twelfth capital M L squared, times thirty six m squared v squared over capital M squared L squared, which gives three m squared v squared over capital M, right? Add them: four m squared v squared over capital M, in total. Now cancel m v squared from both sides. One equals four m over capital M, so capital M equals four m. Capital M over m is four! Three laws, one answer. Got it?

7:48Quick revision: hinged body, free body, elastic or not

So, quick revision. Hinged body: conserve angular momentum about the hinge, because the hinge force has no torque about it. Free body: linear momentum, and angular momentum about the centre of mass. And kinetic energy, only if the collision is elastic. Now go and try the practice problems on this page. All the best!