Mistakes to avoid on this classic rotational mechanics question | IIT JEE Advanced | Irodov

[00:00:00]Why do students keep getting this wrong?

[00:00:02]This is a classical problem from Iridor which will test your concepts on rotational mechanics and once you understand them, you will never get this wrong. Let's read this question carefully to see how best to solve [music] this conceptually. All right, the question says that there is a cylinder of radius R on the left hand side and a small disc here which is tied to the cylinder with a string of length L0 [music] and we are seeing the top view of this setup and then this disc is given a velocity V 0 perpendicular to this string and then we are required to find the time it takes to travel in this path and hits somewhere on this cylinder. Before we proceed, I strongly recommend that you pause the video here and spend some time with it. Once you have done that, unpause.

[00:00:49]Now, before we jump into writing equations, let's pause here to visualize key concepts. Okay. So, here you see the setup from the top view. And what's happening here is that if you play this animation, you would see that the velocity given perpendicular to the string is making it move. And then the string gets wrapped onto the the blue cylinder, right? And then it keeps on moving the path and then it eventually collides with the cylinder. You would see that the initial velocity which is given here is perpendicular to the string. Now there would be a tension that starts acting. However, that will be at 90° to the velocity and the only forces here is this tension. There is no friction which is given in the problem.

[00:01:30]So which means that the velocity at every instant in this motion will remain constant. This is where many students would make a mistake that they would assume that you know about uh the center of the disc will the angular momentum be conserved. Can we say something like mv0 * this uh l0 length is equal to the final velocity at the time of collision or just before the time of collision multiply by mass and the final radius of the cylinder. It is not the case because the forces on the string the tension is acting as a torque about the center of the cylinder. So it's going to change the angular momentum and we cannot conserve that. So if you look at that setup you see that the velocity which is given v 0 this would be remaining the same as is passing through the entire path and then it will eventually collide. The key here is to understand first the velocity is not going to change with time and second that this could be treated as a circular motion.

[00:02:28]However, with a key consideration that at any given point of time, the disc is in a circular position about the point of contact with the cylinder. So, the point at which the thread ends up wrapping, that point is the center of this rotation. And you would see as time passes, there is a variable radi of the circular motion. So, it is not in a circular motion with respect to the center of the cylinder. So with that concept in place, let's try to solve this step by step. To solve this problem, let's use the handy diagram I have on the bottom left of the page where I'm showing the initial position with this yellow lines. This is where the disc was initially having been imparted a velocity of v 0 at an intermediate position before collision.

[00:03:17]I'm just denoting that with uh the [music] blue line. So you see that the the certain length of the string has been wrapped which is over here.

[00:03:25]And then there's a certain free length of the string. Like we discussed earlier, the tension over here is going to keep it stretched out. And the velocity at this [music] point is going to be the same as V 0 which was a key concept students get wrong because again the tension is acting perpendicular to the velocity. And there are no other forces. At the instant of collision which is denoted by the red color over here we see that the entire length L0 of the string would have wrapped up over here and then the disc would have collided with the cylinder. Now at any given instantaneous position say at this point suppose that end point of the string you know after having been wrapped up is subending an angle theta okay with the vertical line and after an infinite decimal small period of time it moves by d theta. Now here again in if you want to depict that on the top you would see that the radius line is moving by d theta then the angle subended over here is also d theta. If we have this angle as d theta we also know that the velocity at this point is again going to be v 0 that is not changing with time.

[00:04:32]Now if we were to just solve this equation where we say that at any given instant the instantaneous velocity is going to be V is equal to instantaneous radius of that circular motion times the angular velocity. Now note that we already understood that V is equal to V 0 at all times. However this radius is not the radius of the cylinder. So this is the instantaneous radius which means we are trying to find out this particular length. That length can be easily understood now as the total length minus the length that is wrapped on the cylinder. And what is the length wrapped on the cylinder? This particular thing is nothing but r theta. Right? So we can now say that the instantaneous length or the radius of that circular motion will be nothing but l0 minus capital r * theta. So we now know the velocity as v 0 r as expressed as a function of theta.

[00:05:27]So can we express omega as a function of theta? Yes, we can. Omega is nothing but the angular velocity which is the rate of change of theta. So if you start putting this value, we say that d theta* dt which is omega is nothing but v / r where v is v 0 divided by the instantaneous radius which is l0 minus r theta. Now with that clarity in place, this equation becomes fairly simpler to solve. All we do is take the theta terms to the left. So we say l 0 minus r theta * d theta is equal to v 0 * dt. Given this equation, we should be able to solve it from the initial instant with time t was 0 where the velocity was imparted to say the final time t of collision where theta changes from zero to certain angle alpha shown here in the circle where the collision happens. And what is alpha? We know that at the time of collision r * alpha is going to be this total length from here to here which will be nothing but the length of the string which should be wrapping up completely. So this will be equal to l0 which implies that alpha is nothing but l0 by r. So if you integrate this equation over here we would simply get L0 theta minus r theta^² by 2 limits varying from 0 to alpha which is L0 by R and the left hand side would be nothing but equal to V 0 * T. Okay, let's put the value over here. So this term gives me L0 * L0 by R which is L0² by R minus L0² by 2 R which is equal to V 0 * T or this simply is L0 2 by 2 R being equal to V 0 * T which simply implies that time is L0² by 2 V R and this is the answer for the time of collision. Now that we know how to solve this question correctly, let's extend this a bit further and try a new variation. In fact, it should be very easy by now to [music] visualize how the tension in the spring changes with rotation. Let me know down in the comments below what that tension would be at any given instant of time before the collision. Great, that's it for this video. Let me know down in [music] the comments below if you have any questions or comments. Keep working on your concepts and I'll see you on the next one. Thanks for watching. Bye-bye.