Hey guys, so in this video, we're going to talk about the rotational dynamics of rolling motion. Rotational dynamics just means you have a problem of torque and acceleration, and rolling motion just means you have an object that rolls around itself while moving sideways. Right? While the axis of rotation moves, while the object rotates. Kind of like toilet paper that instead of being fixed on the wall, you just rolled it on the floor. Right? So it does this. That's what rolling motion is. So let's check it out.

Alright. So in some problems, it says, a disk-like object accelerates around a free axis. Again, free axis is the toilet paper that instead of being fixed on the wall, it's free to move. So it rolls while moving. It says here, the objects will have obviously both rotational and linear motion. For example, in this case, the sky is falling down here. It's going to have, if you release it from rest, it's going to have an acceleration this way, and it's going to have an α this way. Okay. So we're going to use both sum of all forces equals ma because it has an a, and sum of all torques equals Iα because it has an α. And we're going to do this for the same objects. Okay. That same object has 2 motions, so we can write 2 equations. And remember, the direction of positive in both will follow the direction of a, α. Okay? So this is going to be positive and this is going to be positive. Remember also that in rolling motion, rolling motion is the situation we're talking about, there's an extra equation, which is that the velocity at the center of mass, which is this, v_{cm} = rω. Okay? A lot of times you see this equation an equation similar to this. It's v = rω. This one is v = Rω because it always has to do with the radius of this object. Okay? So this is an extra equation that's possible, that's possible because you have rolling motion. Another equation that's also possible is just like a v = Rω, you have a = Rα. Okay. And in these cases, it's always going in this problem in rolling motion, it's always going to be R. Alright?

Now, I want to remind you of some quick things. When you have a disk-like object that is rolling freely, the torque will come from static friction. Okay. Let's talk about that real quick. Which torque are we talking about? Well, this guy has an α. And remember, to have an α, you have to have a torque. Acceleration comes from a net force and an α comes from a net torque. Okay. A torque is what's causing an α. And in this case, the force that's causing this torque, the torque comes from a force. And in this case, the force is static friction. Okay. So in static friction, if the torque is this way I'm sorry, if α is this way, then torque has to be this way which means that friction has to act in a direction that causes this, which would be like this. I hope you can see friction static right there. I hope you can see that if you have a disc and you push your surface, right, and you push in the disc like this, right, I'm doing basically like this. Exactly what friction is doing. So friction goes this way and causes a torque, So friction goes this way and causes a torque of friction. So the force of friction produces a torque of friction, which causes an acceleration. So the disc not only falls this way, but it also does this as it's falling. Okay?

So two things to remember. If there is an acceleration, in other words, if your α is not 0, there has to be static friction. And I like to remember this by using the following short phrase. You need static friction to α. If you don't have static friction, you don't have an α. What that means is that if you're not spinning, you're not going to start. If you're if you're initially not spinning, you have no way of beginning to spin and if you are already spinning, you have no way of stopping your spinning. You will keep spinning because there's not going to be any change. Another keyword is if without slipping. This means that there is no kinetic friction. And guess what? This is always going to be the case. Always. Okay? This is standard language, not slipping. So you're not going to have kinetic friction in these problems. And if you have acceleration if you have acceleration, if acceleration is not 0, okay, if your acceleration is not 0, then there has to be static friction because you need static friction to have acceleration. Cool? Let's do a problem.

So it says here, when a solid cylinder, of mass m and radius r is released from rest, it rolls down without slipping along an inclined plane. Let's draw that. I got an inclined plane like this, just like the picture up above, and I release this from rest. So the initial velocity is 0. This guy has mass m and radius r, and he rolls down. Okay. Rolls down means it's going to do it's going to have an acceleration this way without slipping. Without slipping is standard language, but it does tell you that there is no kinetic friction for sure along an inclined plane that makes an angle of θ. And we want to derive an expression for the angular acceleration of the disc. We want to know what is α. Okay. How do we do this? Well, this is an acceleration problem. So we would use f = ma, except that there's also rotation. So we're going to use f = ma and torque = Iα. And both of them are for the same object because the object has two motions. One object with 2 motions means 2 equations. Sum of all forces equals ma and sum of all torques equals Iα. Okay?

Now let's look at the forces, on this on this thing here. So I'm going to draw it over here. I have an mg pulling you down. I have to decompose the mg into mgx and mgy this way. I have a normal like this and I have a friction that acts over here, friction static. Those are all the forces. In terms of torques, the only torque comes from this. Okay. Normal on a disc that spins on a surface never produces any torque. And mg in the center of mass never produces any torque, if you remember. Right? So the only one here is torque of static friction. Cool? Remember also that once you split mg into mgx and mgy, mg is essentially dead. You got rid of it and replaced it with 2 other things. So it's really no longer there. Remember also that mgy cancels out with normal. So really the only things you have is these are the two forces in the x-axis, which is down the plane, and this is the only torque. Okay. So the forces are, mgx and friction. Now which one is positive and which one is negative? Well, this is the direction of positive. So, mgx is the positive one and friction is negative. That equals MA. Torque, there's only one torque which is the torque of static friction. The moment of inertia, I, is going to be the moment of inertia of a solid cylinder because it says right there, half mr^{2}. In α, remember, we want to replace α with a. If you have a and α, you want to get rid of α and change it into a. And that's so that instead of having a and α, you have, excuse me, you have a and a. To do this, we use the fact that a = rα_{big r} because this is rolling motion. Rolling motion. So I can rewrite α as a / big r. And that's what I'm going to do. I'm going to put a / big r over here and and that's so that this α becomes an a. So I have a and a. Cool? Last step is I have 2 there's not much else to do here. I'm going to expand this one more time. And this is going to be torque of any force f is fr sine of θ. The force we're talking about here, static friction, r is the r vector. In this case, it's always going to be the radius as well because notice that friction always acts at a distance of the entire radius. If you draw an r vector here, r vector, of course, is from the axis of rotation to the point where the force happens. For friction, it's always going to be big R. So the torque of friction is always friction, big R, and the angle is always 90 degrees. So this is always the case for I'm going to put here always. The torque due to friction is always F big R. Okay? So now that I did that on the side, I'm just going to plug it in here. And then look what happens. All the rs will cancel, which is neat. R cancels with this, this cancels with this, and you're left with friction static equals half m a. Now I want to warn you not to get too excited with the expanding of these equations and rewrite friction into new normal. You could do this. You may remember friction is new normal. You should remember friction is new normal. But you don't want to rewrite, you don't want to expand. And that's because, guess what? There's a friction here and there's a friction here. All friction is doing, similar to what you would see with a tension if you had a cable pulling on a block. All it's doing is connecting these two equations. Right? So don't expand friction. Just plug it in right there. So let's do that. Mg, if you remember, Mg is Mg sine of θ. That's mgx becomes Mg sine of θ minus half ma equals ma. Notice here that I wrote little m, little m and big m. And that's because I usually write f = ma with a little m. But I usually write I with the big m because a lot of the equations are like that. It's really the same thing because it's referring to the same object, so I can cancel. If you have a single object, the mass will cancel. Okay. And we're solving for a. So I'm going to move this over here. I have g sine of θ equals a + , half a. K? a + half a is 1 plus half, 1a half, 1.5 or 3 over 2 a. So a is the 2 goes up here, 2g sine of θ, and the 3 goes down there. Boom. Okay. This is the final answer for a. We still got to get α, but I want to again make a point here. Notice how this equation looks familiar. It should look familiar if you have a block that is accelerating down and there's no friction, right, there's no kinetic friction, it's accelerating down because of mgx. The acceleration of this block is g sine of θ. You may remember this from like back in the day, right? G sine of θ. Well, notice this equation is very similar except that instead of instead of having like a one here hiding, right, you have some fraction in front of it. And this fraction is always going to be less than 1. The point here is that in rotation questions that are the rotation equivalents of linear questions, right, Like this is just the rotational equivalent of this. That's a lovely arrow. This is just the rotation equivalent of this. So the equation for a here should look similar to this equation. And that's how you know you are on the right track. Provided of course that you vaguely remember this and you can say, hey, that looks familiar. Okay. So I like to make that point a lot because this way you can know that you're generally in the right direction and it gives you a little bit more certainty that you're correct. And another thing that allows you to do is that if this number here is more than 1 for whatever reason, which it would be if you screw up your signs, then you know that you're wrong and you have to go back. There should always be, less than 1. Alright? So let's wrap it up and find α. α is a / r. So it's just going to be 2g sine of θ divided by 3 r. Okay. This is our final insert and we're done. That's it for this one. I hope this made sense. Make sure to know how to do this stuff. Let me know if you have any questions and let's keep going.