**at this instant, what is the magnitude of its angular momentum relative to point o?** This is a topic that many people are looking for. **amritsang.org** is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, ** amritsang.org ** would like to introduce to you **Chapter 11 Rolling, Torque, and Angular Momentum**. Following along are instructions in the video below:

This video. Were gonna take a quick look at chapter 11 and discuss all all of the things that you should have gotten from the reading some old some so anytime we have rolling as translational motion. Were gonna consider objects that only roll smoothly meaning that theres gonna be no slipping with the wheel and the surface that its rolling on and we can say the center of the mass of the object moves in a line parallel to the surface and objects rotate around their center of mass.

So think of a wheel rotating about its center of mass. Which would be in the center of that wheel and the arc length that it rotates through since theres no slipping this arc length. Here is going to be equivalent to the translational distance.

So were gonna call that s and s. Is just theta. Times r.

Theta being the angle through which we rotate and then from that we can find that the velocity of the center of mass is omega times r. Where omega is the angular velocity. Now if we have pure rotational motion.

As in picture a here that means that at the outer rim. The tangential velocity is equal to the angular velocity. Times the radius.

If we have pure translational motion. Then theyre all moving in the same direction and theyre all moving with the same velocity. So the velocity of the center of mass is the same as the tangential velocity.

However in rolling motion the combination of rotational translational. We know the velocity of the center of mass because this is rolling. Were not moving at the point of contact.

Because there is no slipping and the translational velocity at the top is actually two times. The center of velocity or excuse. Me the velocity of the center of mass and so this allows us to answer.

This checkpoint question here that a is the same and b is less than we know that if objects are moving they have energy and if an object is rotating. It also has energy. So that means when something is moving translationally and rotational e.

Then the kinetic energy is going to be the combination of the two separate kinetic energies. Where we have one half. I omega squared and 1 2.

Mv squared. Now the v. Here is going to be the velocity of the center of mass because we look at that translationally how is this object moving through space.

If the wheel accelerates. Its angular speed must be changing and the force has to cause that change and a force to cause that change and not slip is going to be static friction. And so the acceleration of the object.

This is the linear acceleration is going to equal alpha times r. Now. Thats talking about the center of mass so this bike or the car is accelerating translationally equal to alpha times r.

For smooth rolling down a ramp. We have a nice diagram here that shows why we are going to roll. Because our point of contact is going to be down here at p.

But gravity is slightly off of that and so this is going to cause a rotation or this torque is going to cause a rotation and therefore. This is going to begin to rotate and roll down the hill. This equation right here equation.

1110 is a nice new little simplification that we did not use in ap physics. 1. And this allows.

Us when we have something rolling down. The hill. So this is just for something rolling down a hill.

We can find its acceleration down the hill by using this formula. Where we have g sine theta. Divided by 1 plus.

The moment of inertia divided by mr. Squared okay and so this is a little new it kind of figures in the kinetic energy. Both linearly and rotationally something to note here that friction is causing the rotation.

Its part of the the rotation. The fact that the center of mass is over its support and friction is allowing it not to slip. Then we are going to see the torque in the rotation.

Without friction. The object would simply slide down the ramp yo yos are a fun little toy. But they also include a lot of physics and so as a yo yo moves down the string.

Its losing its potential energy. Which we can find by mgh. But it gains rotational and translational energy.

That energy is being split into two different types rotational and translational and so to find the linear acceleration of a yo yo. We can just assume that well this is a yo yo thats going down a ramp. Its just rolling at an angle of 90 degrees.

The difference here is that it rolls on an axle instead of its outer surface. And its slowed by tension rather than friction. And so we can use the results of equation 1110 to calculate the acceleration of yo yo as it falls and we notices when we throw yo yo down or if we just drop a yo yo.

We see that its acceleration. Its much smaller than just dropping an object. And so say we have a hundred and fifty gram yo yo with an inner radius of three millimeters and its moment of inertia is three times 10 to the negative fifth kilogram meter squared.

So we can find the acceleration by plugging into this. Equation then we get a negative 04. M.

S. Squared much much less than baby g. So previously we define torque as the force times.

The lever arm and maybe even threw in sine theta in there if we had an angle between them and we define it only for rotating body and a fixed axis now we redefine it for an individual particle. That moves along any path relative to a fixed point. So it doesnt need to be a circle.

And what that does is that turns torque into a vector and so the direction of this torque can now be determined with the right hand rule. And this is actually using the right hand rule from cross products that we did way back at the beginning of the semester. And so this defined torque as r cross f.

Now. This is important it matters. Which one we cross with which knowing cross products.

Remembering cross products. The order is particular so its the displacement or the the position vector. I should say not just the displacement.

But the the radius vector the distance from our axis of rotation that vector that position vector crossed the force vector and if we know the magnitudes and the angle between and we can simplify. It to our f sine. Theta or we can say its art to pick perpendicular f.

Or the perpendicular r. Times f. Thats basically what we did back in ap physics.

1. But now were going to look at the general case. Where we have two vectors and we can find the torque vector through a cross product.

Take a look at this checkpoint. Pause. The video read over the answers this diagram here shows a few examples of calculating torque and using the right hand rule to find the direction this diagram can be a little confusing so were gonna look at some examples in class that may be able to help you but if you can follow along with this with this diagram and move your hands to see so you can determine the direction of the torque now one thing to keep in mind here anytime.

We want to find the cross product.

Were always looking at the angle. Thats less than 180. So were putting our fingers along our here and were crossing them or were curling them to our force.

And so in order to have with our right hand on this one our thumb has to be into this screen. And so that gives us the direction of the torque. And so if theres linear momentum and an object is rotating then theres going to be angular momentum and were gonna redefine angular momentum and were gonna use l as the vector and it is a vector and the angular momentum is going to equal the cross product of the position vector with the translational momentum.

Vector okay so wed have to look at the momentum of the object that the linear momentum and take the position cross that momentum vector and well. What is this momentum vector. The momentum vector is just m times.

The velocity vector so really what were doing here is were taking m times. The cross product of the position vector with the velocity vector just like with any cross product. We use the right hand rule to determine the direction and we can simplify if we have the angle between the position vector and the velocity vector if we have that angle fee then we can just take our mv sign fee or if we have the perpendicular components.

We can multiply those as well because of the right hand rule. The momentum. The angular momentum is always perpendicular to the plane formed by the position and linear momentum vectors.

If you recall back to when we talked about momentum. We redefine newtons second law. By saying that the net force was equal to the derivative of the momentum with respect to time.

Well now we can do the same thing angular ly and say that the net torque. That vector is equal to the derivative of the angular momentum with respect to time in order for this to work the torque. And the angular momentum must be defined with respect to the same point typically we pick the origin for this and this all stems back to this equation that we had back in momentum.

Checkpoint. Five here is a good one to take a look at to just review a few things conceptually now in those previous slides. These little ells that that i was talking about here.

Were all about a single particle. So this is looking at single particles in a system. So just like with the moment of inertia.

We can sum up all of the little particles mass times. The radius squared to get the total moment of inertia of the object. We can do the same thing with the angular momenta.

If we sum up all of the angular momenta or use this sum. Here. And this is more of a theoretical sum.

Not that were physically going to do this. But just the idea of summing up all of those tiny little momenta. We get the big l.

Momentum. Which is our total angular momentum of an object. And so the derivative of that with respect to time or the rate of change of this net.

Angular momentum is going to be the sum of the net torque on the entire object or in other words. The net work on the entire object. The whole rigid body is going to equal the derivative of the total angular momentum.

And you might think that this is a little repetitive from from overstated before which it is and it is in in that it is the same relationship and thats important that the torque on an object is going to equal the rate of change of the momentum of that object. But now this is talking about instead of just a particle. Its the entire object the entire body itself and so using the same logic.

Im gonna go through this here. Very quickly if we look at the angular momentum and were gonna add up the little tiny momentum moment of each particle. Well we know we define that as the mass times the velocity times the the position vector and if were just looking at the perpendicular portion.

We know that omega times r. Is our v. And then we get an mr.

Squared and the sum of em r. Squared. We know gives us our moment of inertia of the body.

So this simplifies for us to say that okay the angular momentum of an object that is fixed at an axis is just simply going to be i omega. And and it should make a little intuitive sense in that eye well. Thats kind of like a mass and omega is kind of like a velocity.

And so momentum is mass times velocity. So angular momentum is angular mass times. Angular velocity.

Now this table 11 1. Is very important in that it gives us the relationships between our translational objects. Our translational formulas and our rotational formulas just like before when we have momentum.

We know that the law of conservation of momentum exists well that occurs in translational momentum. And it occurs in angular momentum. Now the key thing here is they are conserved separately angular momentum and rotational momentum are conserved separately.

They do not we dont exchange from one to the other and if you think back in ap physics. When we watch that video of the bullet shooting into a block and shooting up into the air that that kind of illustrates that angular momentum and linear momentum are conserved separately. Now when we have the conservation wrangham angular momentum excuse me basically were going to see that the initial momentum is equal to the final momentum.

So i alpha initial is going to equal i ill file and weve seen all of these experiments before of showing the conservation of angular momentum. When we are spinning and we pull our arms in because were changing a moment of inertia. Its changing the rate at which we rotate well this can be seen in a student spinning on a stool.

A springboard diver or a long jumper. All of those conserving their angular momentum by changing the moment of inertia and in turn. The angular velocity has to change take a look at checkpoint.

Seven here talking about our announcer. Rhinoceros beetle. And see if you can get these answers the last thing here in chapter 11 is talking about the per session of a gyroscope and were gonna look at this more conceptually than anything.

But basically a spinning gyroscope in youve seen demonstrations of this youve seen the the spinning wheel in class. Where you can hold it up and and so the the reason that this works is because of the conservation of angular momentum. And so when we have something thats spinning there is angular momentum and we its its very hard to change that and that momentum wants to be conserved.

And only torque can change the angular momentum. So on outside forces is what can change it and without that outside force. Its gonna continue on forever.

And there is this one equation that comes up were not gonna use it a whole lot. But its talking about the precession rate and we use capital omega here and thats when we have the the mass times baby g. Times r.

Divided by i omega so you can see a few things in here. Well. Weve got mg r.

And we have i omega you know the i omega being our momentum. Now this is true for something that is spinning pretty fast if youre spinning slowly then theres not a whole lot of angular momentum. So this doesnt really work all that well now notice this point.

Here. This precession rate is actually independent of mass and you might say well. How is it independent of mass.

Theres m right there in the equation. Well yes. There is theres an m in the numerator.

However i in the denominator is also proportional to m. Its dependent upon m. So those ms hansol now we leave it there just in case.

We have it in terms of em and i depend depending upon the given information depending on you know it if you have the momentum or if you have the moment of inertia so on and so forth. But however the mass does that matter but g does so it does matter what planet were on it just doesnt matter the mass of the object so i know i flew through things pretty quickly there. But i want to try to get all chapter of eleven into one cohesive video and these last two slides.

Here are a nice summary of all of the equations that we would need to have ready for the test. .

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