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## Angular Momentum & Newton's Second Law

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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 first wanna learn Africans. They may can be re written in terms of linear momentum. Instead of Africans, they may. I can write it. Some of all forces equals Delta P change in momentum over Delta T changing time or Time Elapsed, the rotational version of Newton's second Law. Some of all torques equals I Alfa can also be rewritten, but in terms of angular instead of linear momentum because it's the rotational version. So instead of some of those forces, I'm gonna have some of all torques. And instead of Delta P, I'm gonna have delta l l is the equivalence, the rotational version, if you will off p so out and then this is just type. All right, let's do a quick example to see how we would use this equation to solve a problem. So I have a small objects. Small means it's gonna be a point mass. Okay, um, 10 kg in mass mass equals 10 spins within rpm of 80. So our PM equals 1 80 in a circular path of radius five. Now the radius of a circular path is really the distance to the center. That's little R equals five, and you haven't em here. And the object is going this way with some Omega. It says that you apply a constant torque of eighties, so torque equals 80 in trying to stop it. So you were applying If the object is spinning this way, the question doesn't tell you which way it spends. But let's say it's spinning this way. So you're applying a torque in this direction to try to stop it. I wanna 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 zero to an Omega final, that is zero. So we're stopping and I wanna 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 were gonna dio some of all torques equals Delta L over Delta T I'm looking for Delta T, so I'm gonna I'm gonna solve it all the time. Gonna move it up and I'm going to move. So delta l is here on I'm gonna move torque down. There's only one torque here, which is the 80. So I'm gonna 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 or make a final I initial Omega initial divided by 80. Now we are coming to a stop, so make a final will be zero. So this whole thing is zero. So we're left with, um, we're left with, um I initial omega initial mine divided by 80. Okay, um, technically, I guess I should have put this This torque here is causing you to slow down, so it's a negative torque. OK, it's negative torque because it's causing you to slow down. Also the way that I drew it, Um, I have this as a positive because it's clockwise or counterclockwise, and torque is this way. Eso this is going to be negative. It's clockwise and also it's trying to stop you. So I'm gonna 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. Uh, not too big of a deal. So what we're gonna 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 are is five. So this is going to be 2 50. And so I got to 50 there. The negatives cancel. And then omega I have to get Omega's. Well, Omega is two pi f, which is two pi r p m over eso. It's two pi 1 80. Divided by 60 is three, three times to buy. 65 Okay, so six pie goes right here and we are done. You multiply this whole thing, multiply this whole thing and we get seconds. Okay, 59 seconds. So that's how long it would take to cause this object to slow down if you apply this talk. All right. Very straightforward. Just had to plug it into the equation. Three only part that you have to sweat a little bit is that when you expanded, Ally, you had to go find I and then you had to go find Omega. Um, but other than that, it was just straight forward plug into the equation. All right, that's it for this one. Let me know if you have any questions and let's keep going.

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Problem

A solid disc of mass M = 40 kg and radius R = 2 m is free to rotate about a fixed, frictionless, perpendicular axis through its center. You apply a constant, tangential force on the disc’s surface (as shown), to get it to spin. Calculate the magnitude of the force needed to get the disc to 100 rad/s in just one minute.

A

33.3 N

B

66.7 N

C

133 N

D

167 N

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