Hey, guys. So in this video, I want to show you how there's a relationship between angular momentum, l, and the rotational version of Newton's second law. Let's check it out. So remember that Newton's second law, the linear version, the first one you learn, f equals ma, can be rewritten in terms of linear momentum. Instead of f equals ma, I can write the sum of all forces equals delta p, change in momentum over delta t, change in time or time elapsed. The rotational version of Newton's second law, sum of all torques equals I alpha can also be rewritten, but in terms of angular instead of linear momentum because it's the rotational version. So instead of sum of forces, I'm going to have sum of torques. And instead of delta p, I'm going to have delta l. L is the equivalent, the rotational version, if you will, of p, l. And this is just time. Alright. Let's do a quick example to see how we would use this equation to solve a problem.

So I have a small object, small means it's going to be a point mass. Okay. 10 kilograms in mass, mass equals 10, spins with an RPM of 180 so RPM equals 180 in a circular path of radius 5. Now the radius of a circular path is really the distance to the center, so that's little r equals 5 and you have an m here and the object is going this way with some Omega. It says that you apply a constant torque of 80, so torque equals 80. In trying to stop it, so you are applying if the object, is spinning this way, the question doesn't tell you which way it spins. But let's say it's spinning this way. So you're applying a torque in this direction to try to stop it. I want to know how long will it take for it to come to a complete stop. So I'm going from some Omega initial that is not 0 to an Omega final that is 0. So we're stopping and I want to know how long does it take to do that. Now notice that I'm giving you torque and I'm asking for delta t. Therefore, this equation right here can be used, right, because it's got both of those. That's what we're going to do. Sum of torques equals delta l over delta t. I'm looking for delta t. So I'm going to, I'm going to solve for delta t. I'm going to move it up and I'm going to move so delta l is here and I'm going to move torque down. There's only one torque here which is the 80. So, I'm going to put that in here. Delta l is l final minus l initial and then this torque is 80. L, remember, is I omega. So you can think of this as I, final omega final, I initial omega initial divided by 80. Now we are coming to a stop, so Omega final will be 0. So this whole thing is 0. So we're left with, we're left with, I initial, omega initial, minus, divided by 80. Okay. Technically, I guess I should have put this torque here is causing you to slow down. So it's a negative torque. Okay? It's a negative torque because it's causing you to slow down. Also, the way that I drew it, I have this as a positive because it's clockwise or counterclockwise and torque is this way. So this is going to be negative. It's clockwise and also it's trying to stop you. So I'm going to put a negative here. And I needed that negative to cancel out, otherwise, end up with a negative time. Okay. But that's just a sign thing, not too big of a deal. So what we're going to do now is plug in these numbers. I don't have I, but I can find it because I is going to be the I of a point mass, which is m r squared. M is 10. R is 5. So this is going to be 250. And so I got 250 there. The negatives cancel. And then omega, I have to get omega as well. Omega is 2πf which is 2πrpm over 60. So it's 2 π, 180 divided by 60 is 3. 3 times 2π, 6π. Okay. So 6π goes right here, and we are done. You multiply this whole thing multiply this whole thing, and we get 59 seconds. Okay. 59 seconds. So that's how long it would take, to cause this object to slow down if you apply this torque. Alright? Very straightforward. Just had to plug it into the equation. The only part that you had to sweat a little bit is that, when you expanded I, you had to go find I and then you had to go find omega. But other than that, it was just straightforward plug into the equation. Alright. That's it for this one. Let me know if you have any questions and let's keep going.