Concept: Angular Momentum & Newton's Second Law6m
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 one you learnt, F equals ma, can be written in terms of linear momentum instead of F equals ma I can write the sum of the forces equals Delta P, change in momentum, over delta T, changing 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 all forces I'm gonna F sum of all torques and instead of delta p, I'm going to have Delta L, L is the equivalence the rotational version if you will of P so 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, 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 we're gonna do, sum of all torques equals Delta L over delta T, I'm looking for delta T. So, I'm going to, I'm going to solve for that delta and 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 divided by 80, okay? Technically I guess I should have put this, this torque here is causing it to slow down. So it's a negative torque, okay? 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 in this way. So this is going to be negative, its clockwise and also it's trying to stop you. So I'm going to put a negative here and I need of 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 got 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 mr 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 2pi, F which is 2pi, rpm over 60. So it's 2 pi, 180 divided by 60 is 3, 3 times 2pi, 6pi, okay? So 6 pi 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 have to plug it into the equation, the only part that you have to sweat a little bit is when you expand Li you have to go find I and you have to go find Omega but other than that it was straight forward plug into the equation. Alright, that's it for this one, let me know if you have any questions and let's keep going.
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.6m
A disk has a moment of inertia of 3.0 X 10-4 kg•m2 and rotates with an angular speed of 3.5 rad/sec. What net torque must be applied to bring it to rest within 3 s?
a. 4.5 x 10-3 N•m
b. 7.5 x 10-4 N•m
c. 3.5 x 10-4 N•m
d. 5.0 x 10-4 N•m
A car's wheels will typically be 10 kg in mass and 45 cm in diameter. If a car is moving at 30 m/s and can brake to a full stop in 5 seconds, how much torque do the brakes apply to the wheel?
A solid disk of mass 15 kg and radius 10 cm starts at rest, undergoes some toruqe, and is rotating at 5 rad/s after 0.5 s. (a) What is the change in the disk's angular momentum? (b) What is the torque applied to the disk?