Subjects

Sections | |||
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Intro to Angular Momentum | 15 mins | 0 completed | Learn |

Jumping Into/Out of Moving Disc | 24 mins | 0 completed | Learn |

Opening/Closing Arms on Rotating Stool | 18 mins | 0 completed | Learn |

Spinning on String of Variable Length | 20 mins | 0 completed | Learn |

Conservation of Angular Momentum | 46 mins | 0 completed | Learn |

Angular Collisions with Linear Motion | 8 mins | 0 completed | Learn |

Angular Momentum & Newton's Second Law | 11 mins | 0 completed | Learn |

Intro to Angular Collisions | 15 mins | 0 completed | Learn |

Angular Momentum of a Point Mass | 22 mins | 0 completed | Learn |

Angular Momentum of Objects in Linear Motion | 7 mins | 0 completed | Learn |

Additional Practice |
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!! Conservation of Angular Momentum with Energy |

Concept #1: Angular Momentum of Objects in Linear Motion

**Transcript**

Hey what's up guys. So in this video we're going to talk about calculating the angular momentum L of an object that has linear motion. Now remember angular momentum is the momentum of an object that rotates. So it might seem weird they would find the angular momentum of an object that is moving in a straight line that is not rotating, let's check this out. So says here, in some problems an object moving in a straight line will collide against an object that is fixed in a rotating axis. So here, for example, I have these sort of like a rotating door, that's fixed about this axis this object here will collide here and it will cause this door to spin this way let's say, right? So it is in a situation like this that the angular momentum of this guy. So the L of that object will make sense to be used, okay? So you may remember that, we use linear momentum to solve collision problems, well, to solve linear collision problems. So if you have a problem where both objects have linear velocity as they collide then we're going to solve this, we're going to solve with linear momentum, with conservation of linear momentum but if we have a situation where one object has a V and another object has an Omega like here, the subject moves with a V but then after the collision the doors will have an Omega then, we're not going to use p to do this, we're going to use L, okay? So I'm going to write here, we're going to solve this with L. Now if I have two objects with Omega. So if you have two disks and they're both spinning and you push them against each other that is an angular collision, I have an Omega meeting up with another Omega and therefore we're going to use not linear momentum but angular momentum, okay? These two here are probably obvious if you have two linear motions you use linear momentum if you have two an angular momentum, two angular motions you use angular momentum, what's interesting is the middle one which is when you have one of each L takes over. So what matters is the L not the P, we're going to do some collision questions later and you see this happen. So when, we're trying to figure this out in this case here, we need to first find the object's angular momentum L and not its linear momentum because of what I just said here, these questions will be solved with L. So I don't care what this guy's p is, I care what L is, okay? But the question is, how do we get the angular momentum, how to get an angular momentum of an object that's moving in a straight line? This object isn't even rotating, how do I find its rotating momentum? Well, an object in straight line has angular momentum relative to an unrelated axis of rotation, what do I mean? is it can actually find the angular momentum of this guy relative to this axis even though they haven't really collided yet, right? So it's an unrelated axis. So hey that axis over there, let's find an angular momentum relative to it and we use the equation L equals mvr, okay? And notice that this is the same equation that we used for the angular momentum of a point mass, angular momentum of a point mass, okay? So, let's do an example and see what the deal is here. So two rotating doors each six meters long are fixed to the same central axis of rotation as shown above, this is a top view which means you are looking down into the doors and you see them from the top, okay? So, we got those two there and I'm supposing bird, 4 kilogram bird. So the mass of this guy here 4 kilograms, the bird moves of the velocity of 20, 30 actually, plus thirty horizontal. So it's about to collide against the door. So your top view you see the bird going like this, is about to collide against the door at a 0.50 meters from one end. Now it says here that it has two doors which means we're talking about one, two and the doors are 6 meters long, okay? Which means each one of these points here is 3 meters long. So you can think of this as a half a door, okay? So it means that this whole thing here is 3 meters. Now the bird is colliding at 50 centimeters from one end. So the bird is colliding 50 centimeters from one end obviously, we're talking about this end and not this end, okay? So this is 0.5 meters which means that this distance here is 2.5 meters 0.5 and 2.5 okay, we want to know the birds angular momentum about the access through the center of the door just before hitting the door. So, again, the angular momentum of an object in linear motion is given by L equals mvr mass is 4, V is 30, those are just plug, just plug into the equation and r is the distance from the axis of rotation to the point where it will touch, okay? So just like a bunch of the other little Rs that we derived, that we've used, it's just a distance from the axis of rotation to the point where the point of interest which is a point of collision, that distance is 2.5, okay? So this is very straightforward this is 0.5 so we're going to use this distance here because it collides here. So it's 2.5. So 2.5 goes right there, we multiply this whole thing and we have that this is 300 kilograms meters per meters squared per second okay, that's it just straightforward plug it in here, just warning you that this is going to come back later, it's going to make a return when we fully solve these problems, these types of rotational collisions. So later on this is going to collide, the doors are going to spin and we're going to actually be able to calculate how fast the door spins but not yet. So contain your excitement and let's keep going, let me know if you have any questions.

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Concept #1: Angular Momentum of Objects in Linear Motion

A particle of mass m near the surface of Earth is launched with an initial velocity v0 at an angle θ above the horizontal. Using the origin as the pivot, find the angular momentum when the particle is at the highest point of its trajectory. The z-axis points out of the page.
1. L = mv03/2g sinθ cos2 θ k
2. L = mv03/2g sin2 θ cosθ k
3. L = mv02/2g sin2 θ cosθ k
4. L = - mv02/2g sinθ cos2 θ k
5. L = - mv03/2g sin2 θ cosθ k
6. L = mv02/2g sinθ cosθ k
7. L = - mv02/2g sin2 θ cosθ k
8. L = - mv02/2g tan3 θ k
9. L = mv02/2g tan3 θ k
10. L = - mv02/2g sin θ cos2 θ k

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure (Figure 1). At this instant, what are the magnitude and direction of its angular momentum relative to point O?

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure (Figure 1). If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?

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