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Concept #1: Intro to Angular Collisions (Two discs)

Transcript

Hey guys. So in this video we're going to start talking about angular collision and angular collision happens when you have either two disks that are rotating and if you like push them against each other so that they now spin together, so that's an angular collision or if you have an object moving in linear motion that hits another object that will rotate. So, for example, if you have a bar and let's say some object comes and hits the bar the bar is fixed here. So the bar will rotate, okay? That would be an angular collision problem resolved. So let's check it out. So angular collision happens, angular collisions will happen when one of the two objects, at least one. So I should say one or more of the two objects either is rotating or rotates as a result, the problem I mention here, this hits, this object will rotate as a result. So this is an angular collision, okay? So there are actually three different setups of collisions, you can have a collision where two objects have linear motion like, for example, two boxes going towards each other, in this case this is a linear collision and we're going to use, this is linear collision, we're going to use conservation of linear momentum P to solve this problem. If you have two objects that are rotating it should be obvious that this is an angular collision or rotational collision. So the example I gave two disks that are spinning and you push them together so they spin together, and we're going to use this as an angular collision, we're going to use conservation of angular momentum to solve this. So the non obvious case is if you have one V and 1 Omega which is the example I gave with the bar, right? So this object moves at the V hits the bar here causes the bar to spin the object has V the bar will get an Omega as a result. So we use this to solve this, this is action angular collision, even though there's one of each it's not a linear collision, it's considered an angular collision and we're going to solve it using the conservation of angular momentum L, you can think of it as L basically supercedes as long as you have one Omega L will take over for p, okay? So this I already mentioned briefly, similar to linear collision we're going to use the conservation of momentum equations but we're going to use obviously the angular version, okay? That's what this is, we're going to use conservation of L instead of conservation of p. So instead of P initial equals P final I'm gonna write L initial equals L final, remember the conservation of momentum equation for linear momentum if you expand the whole equation you would have m1v1 m2v2, m1v1 ,m2v2 it's the same thing here but mv will be replaced by I Omega. So it's going to I1 Omega1 initial plus I2 Omega2 initial equals I1 Omega1 final I2 Omega2 final, cool? That's the conservation of angular momentum equation. Now if you have a point mass in linear motion we're going to use the linear version of the angular momentum equation, what is this? So if you have the situation I keep describing, if you have an object that collides against the bar. So you have a mass M here moving this way with a velocity V and he hits the bar at a distance it's the bar right there at a distance little R from the axis of rotation, we're going to use the equation that L equals mvr, let me put this over here as well L equals mvr, okay? I guess it might make more sense to put this over here on the other side that, so these guys are hanging out together, cool? L equals mvr, is what we're going to use. Now what that means is that for that object instead of using I Omega, instead of using I Omega you're going to use this. So I'm going to write here instead, okay? We're going to do this don't worry we'll do an example. Now in this equation r as I mentioned here r is the distance between where the linear object collides, right here red dot, and the the axis of the rotating object the blue dot, right? So it's just the r vector between those two points and the last point I want to make here is, before we go to an example, is if you have a situation where you have a rotating disk and you add mass to the disk that is technically an angular collision problem though we could have solved that without talking about angular collisions just by using conservation of angular momentum, okay? And the reason we could do that without worrying about a, you know, different implications of linear collision mixed with angular collisions is that these questions are simpler if the mass was at rest. So you can just add a mass in there, it's a much simpler problem don't worry about it we'll get there as well. Alright, so here I have two disks, the blue disk if you read the whole thing here, the blue disk is spinning, notice I have a disk of radius 6 and a disk of radius 3. So this is the 6 obviously. So I'm going to call this r1 equals 6 meters and it has a mass of 100. So mass 1 equals 100 and then this is, I'm going to call this r2, it's 3 meters and it has a mass m r2 m2 of 50, okay? It says here that's the 100 kilogram, so the outer one, the bigger one spins clockwise, clockwise looks like this at 120 rpm. So I'm going to say that the rpm is 120, I'm going to call this negative 120 because it's clockwise, clockwise is negative, okay? And spins around a perpendicular axis through its center, what that means is if you have a disk perpendicular axis just means that your imaginary axis line runs 90 degrees to the face of the disk. So it just means that the disk is spinning like this, standard rotation for a disk, okay? A second solid disk which is the darker one there is carefully placed on top of the first disk and it causes the disk to spin together. So imagine one disk is already rotating, the blue disk is already rotating and then you add the gray disk top of it and now the blue disk is essentially carrying the gray disk and they're spinning together, how do you think this is going to affect the speed, the angular speed Omega or the RPM of the blue disk, I hope you're thinking if you add some stuff on top it's going to spin slower and that's what's going to happen, okay? We're going to have a lower rpm. So this question is asking us to find the new rates in rpm that the disks will have in two different situations. So, first we're going to add a disk we're going to add a smaller disk here at rest. So we'll just lower it slowly and in another situation we're going to have it where the disk on top was actually rotating in the opposite direction so now we're going to have a disk spinning in one way and the other disk spinning the other way and we're going to land one disk into the other, okay? So let's do that, on the first one we're saying that the initial Omega of disk 2 is 0 but this one has an initial rpm, this is rpm of 1 of negative 120, there's two ways you can go about this, you have Omega and you have rpm, I'm giving you rpm and I'm asking you for rpm the question here is what is, this is initial, what is RPM final of the whole system, they will rotate together, right? So what is RPM final of the whole system I'm giving an rpm I'm asking for an RPM but remember the momentum equation, the angular momentum equation has omega and not RPM, so you have two choices, you can convert from rpm, you can go from rpm to Omega, do your calculations and then come back to rpm or you can just rewrite the angular momentum equation, the L equation in terms of RPM instead of omega, I'm going to do that instead because I want to show you how that would look, okay? So conservation of angular momentum, you're doing something that changes the rotation of a system. So we're going to start with Li equals Lf, okay? In the beginning all you have is you have the blue disk spinning by itself and the the great disk doesn't spin at all. So all I have is I1 Omega1 initial, at the end they're going to spin together they both have rotation. So I'm going to have I2, if you want you could have written it this way, right? I2, Omega2 initial and just say that this is 0 because that disk is at rest, okay? And then this is going to be I1 Omega1 final plus I2 Omega2 final, I hope you realize that this is going to be the same, okay? Omega1 final equals Omega 2final. So we're just going to call it Omega final because they are, they're going to spin together as a results, okay? You may even remember that these situations where two objects collide and after the collision they move with the same speed is called a completely inelastic collision. So this is technically a completely inelastic angular collision, cool? Fascinating. Alright, so we're going to be able to do this here Omega final I1 plus 2 and this here is just I1 omega1 initial and then what I want to do, we don't have omegas we have we have rpms. So I want to rewrite omegas in terms of RPM. So I1, instead of Omega I'm going to have 2pi, rpm1 initial divided by 60 equals 2pi RPM final divided by 60 and that is times I1 plus I2, okay? I'm going to cancel 2pis, I'm going to cancel the 60s and you end up with this, I'm giving you that this is 120 and I'm asking you for this all you got to do now is plug in I1 and I2, so let's do that real quick. So I'm going to go off to the side here and find I1, I1 is half m1 r1 squared half m1 is 100 the radius is 6 squared and if you do this you get 1,800 for I2 you have a half m2 r2 squared, I'm going to calculate this off to the side because we're going to use these a lot, this is going to be half 50, 3 squared. So that's that's going to be 225, okay? Yep 225. Alright, so let's plug these numbers back in here I1 is going to be 1800 times the initial rpm, the initial rpm is this one right here of the first disk, it's negative 120 equals rpm final which is our target variable and then we're going to add the two Is. So 1800 plus 225, I combine this, I move it over to the other side and I get that the final rpm of the joint system is going to be negative 107. Now this should make sense, the disk was spinning with 120 once you added something to the top of the disk it now slows down a little bit goes from 120 to 107 still spins in the same direction which is a negative direction. So the final rpm is 107, I guess we should say here clockwise. Now that's part a where the disk that we put, the smaller great disk had no initial speed. Now for Part B that disk will have initial speed and we're tight on space here but I'm going to cram it in here and we're going to be fine. So for Part B same equation I1 Omega one initial plus I2 Omega2 initial equals I1 Omega1 final plus I2 Omega2 final, remember I can rewrite Omega as 2pi rpm over 60, okay? That's what we did here. So I'm going to rewrite all of these omegas as 2pi rpm over 60. So I'm going to have every single one of these four terms will have a 2 pi and a 60. So I can cancel the 2 pi and 60 on all of them, essentially I'm replacing W just with rpm because 2 pi and 60 will cancel everywhere I1 is the same here 1807, right? 1800 rpm1 initial plus 225 rpm2 initial equals, the Omega here is the same because they spin together and then I just add up, and I'm also going to rewrite this as rpm as well. So rpm final of both, this is our target variable and I'm going to add up the Is. So it's going to be 1800 plus 225, okay? The RPM of the first one in this problem, the bigger disk rotates with 120 clockwise so this is going to be negative 120 and it's saying here that the second disk for Part B would be spinning counterclockwise so positive with an RPM of 360. So here this would be plus 360, okay? So if you multiply all this crap on the left you've got a big negative number a smaller positive number and if you combine those the left side combines to be negative 135 thousand and the right side is 2025. So rpm final times 2025. So I'm going to move the 2025 to the other side and divide the two and we get to the final answer which is negative 67. So the final rpm is negative 67, this means that it's also clockwise, let's talk about this real quick what this means is. So this is the second answer here for Part B, very similar set up to part a, just the only thing I changed was this here that instead of 0 was 360, okay? So it let's talk about this real quick, it was speeding at 120 if you add in a disk that doesn't spin at all it just makes it heavier. So it's going to go from 120 to 107 but if instead you get a disk that's staying in the other direction, right? So one disk spinning clockwise and then you add a disk that is counterclockwise the final of the two will be lower and that's because the bigger disk has to slow down a lot to cancel out the opposite rotation of the other object. Alright, so this, hopefully this makes sense but let me know if you have any questions and let's get going.

Two disks of identical mass but different radii (r and 2 r) are spinning on frictionless bearings at the same angular speed ω0 but in opposite directions. The two disks are brought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity. What is the magnitude of that final angular velocity in terms of ω0? 1. ωf = 2/3 ω0 2. ωf = 1/5 ω0 3. ωf = 1/3 ω0 4. ωf = 1/2 ω0 5. ωf = 4/5 ω0 6. ωf = 3/5 ω0 7. ωf = 3/4 ω0 8. ωf = 2/5 ω0 9. ωf = 1/4 ω0
A stationary bicycle wheel of radius R is mounted in the vertical plane on a horizontal low friction axle. Initially the wheel is not rotating. The wheel has mass M, all concentrated in the rim (spokes have negligible mass). A lump of clay with mass m falls and sticks to the outer edge of the wheel at an angle θ as shown in the figure. Just before impact the clay has a speed v. Just after impact, what is the magnitude of the angular velocity of the wheel? 1. mvcosθ / MR 2. mv / (m + M)R 3. mv / MR 4. mvcosθ / (m + M)R 5. mvsinθ / (m + M)R 6. mv / (m + M)R 2 7. Mv / mR 8. Mvsinθ / mR 9. mv / MR 2 10. mvsinθ / MR
A disc with moment of inertia I1 = 40 kg • m2 and angular velocity ω1 = 20 rad/s is dropped on to a stationary second disc along the axis of rotation. The second disc has moment of inertia I2 = 60 kg • m2. How much rotational kinetic energy is lost? A. 3200 J B. 11300 J C. 8000 J D. 15000 J E. 4800 J
A disc with moment of inertia I1 = 40 kg • m2 and angular velocity ω1 = 20 rad/s is dropped on to a stationary second disc along the axis of rotation. The second disc has moment of inertia I2 = 60 kg • m2. What is the angular velocity of the two discs? A. 12 rad/s B. 8 rad/s C. 4 rad/s D. 20 rad/s E. 6 rad/s
A thin uniform cylindrical turntable of radius 3 m and mass 25 kg rotates in a horizontal plane with an initial angular speed of 7.9 rad/s. The turntable bearing is frictionless. A clump of clay of mass 12 kg is dropped onto the turntable and sticks at a point 1.1 m from the point of rotation. Find the angular speed of the clay and turntable. A. 7.75532 B. 5.75699 C. 6.99693 D. 11.0707 E. 7.097 F. 8.15289 G. 8.35556 H. 4.96894 I. 8.6087 J. 9.56476
A sticky ball collides with and sticks to rod that is floating free in space. Consider the system of the ball and the rod, and let E be the kinetic energy of the system, p the linear momentum of the system, and L the angular momentum of the system. List which of these quantities are conserved about the point where the putty strikes the rod.1. p; E2. L3. L; E4. L; p; E5. E6. p7. L; p8. none