Ball hits rod angular momentum example | Physics | Khan Academy
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- Опубликовано: 21 июл 2024
- In this video David explains how a mass can have angular momentum even if it is traveling along a straight line. Then David shows how to solve the conservation of angular momentum problem where a ball hits a rod which can rotate. Created by David SantoPietro.
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I enjoy ur explanation so much thx.very thx.
Amazing !!!💜💚✌✌💗
*Please correct me if I'm wrong, I'm still new to these concepts*
I was incredibly confused with what appeared to be a new formula for angular momentum when I had a thought: The angular momentum of the (disconnected) ball here is actually the same as the angular momentum of the ball if the ball were connected physically to the axis at the point of collision - its just a different way of writing the formula. The angular momentum formula that we are familiar with is L = Iw. Because the ball is acting like a point mass, I = mr^2. We get L = mr^2w. Since tangential velocity (v) = wr, we can just substitute that in and we get L = mrv. r is the "radius", which in this case, is the distance between the ball and the axis at the point of collision and v is the speed at which the ball is travelling (if you look at the direction of the ball, it is travelling perpendicular to the bar, which is why its velocity is tangential velocity relative to the axis). So this wasn't a new formula, it was just a matter of perspective :D
Should the moment of Inertia about an end of a rod not be 4/3mr^2?
Dam. That's one long hard rod
💀
Could you use conservation of kinetic energy to solve this problem as well?
So what if the ball comes in contact with the rod at an angle? Does this formula still hold? I imagine you would need to take into account only the component of speed that is perpendicular to the radius.
Multiply sin (angle) with the velocity of the ball is hitting at an angle
V_perpenpicular = V * sin(angle)
And yeah this video is actually covering the balls angular momentum. Hitting the rod is used here as an intution. The main focus of this video is to determine the angular momentum of a object moving in a straight line.
Would this setup work as a reactionless drive? It seems that if a ball Would bounce back and forth from a spinning rod to a wall of a box in which said spinning rod is contained ,the reaction from the box wall would be absorbed as angular momentum, thus resulting in a net liniar momentum of a sistem. I have determined experimentally that this is not the case, but I cant figure out the fault in my reasoning. If anyone could help me understand how liniar momentum is conserved I would be very gratefull.
I would think it could be visualized on the way that the moving object is rotating around the centre of the earth which is too much far way; consequently, the inertia of the rotating ball is high enough to make its angular velocity is slight.
What if the ball was sticky and stuck to the rod?
Why don't we apply law of conservation of energy?
That's my question
Isn't it true that angular momentum is always calculated about a point not about any axis. Though in this case math would remain same, but this is something to be taken care of.
There is something i don't understand here. Considering all collisions are elastic, if the ball (a point particle in this case) with a mass of 5kg hits the rod (at rest) of mass 10kg at a speed of 8m/s, how can it stop, or bounce back at the same time ? You didn't change the initial variables, so either it bounces back, stops, or continues with less speed. But saying it can either do that or this without changing the values seems odd.
I saw your comment and decided to try to use conservation of kinetic energy to find which velocity outcome would happen.
The video just used conservation of angular momentum: Li = Lf
He chose a final velocity but if we leave it as a variable v:
0 + mvᵢR = Iω + mvR
ω = mR(vᵢ - v)/I
We can't know ω or v from this alone.
But if it was elastic, we can use conservation of kinetic energy: Ki = Kf
0 + ½mvᵢ² = ½Iω² + ½mv²
ω² = m(vᵢ² - v²)/I
We can plug ω in from the momentum equation:
(mR(vᵢ - v)/I)² = m(vᵢ² - v²)/I
m²R²(vᵢ² - 2vᵢv + v²)/I² = m(vᵢ² - v²)/I
This rearranges to a quadratic equation in v:
(mR²+I)v² + (-2mR²vᵢ)v + (mR²-I)vᵢ² = 0
The solutions are v = (mR²vᵢ ± I)/(mR²+I)
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Except... there's some things I don't understand either.
First, there's a force applied on the rod at the axis, that holds the left end in place.
The video addressed that this doesn't affect the conservation of angular momentum (0 torque).
But does this take energy out of the system?
Second, is there any way the ball could go right or left after hitting the rod? the rod does start heading left as it turns, so I wondered about that; maybe the ball should go right to cancel this.
Third, I don't know how to tell which of the two solutions for final velocity is the one that happens.
If we plug in the numbers, they come out to 5.2 m/s or 4.4 m/s.
The angular velocity for the rod is then 1.05 rad/s or 1.35 rad/s.
The velocity of the right end of the rod upwards would be 4.2 m/s or 5.4 m/s.
So I guess if v = 5.2 m/s it'd collide with the rod again. If v = 4.4 m/s the rod would be able to get out of the way (depending on the radius of the ball), so maybe that means 4.4 m/s is the right answer but I'm not sure personally.
I want to know why angular momentum is conserved since the rod does experience a torque at the end where ball hits it
Kunal Verma angular momentum is conserved because there is no external torque applied to the system. It's true that if you took the system to be the rod alone, then there would be a change in momentum due to the torque from the ball, but here we are considering the ball and the rod as the system. In this situation, the torque is just what transfers the momentum from the ball to the rod and is equal in opposite directions. So since the ball stops and the rod moves, the net change in momentum is 0
can't regular momentum convert to angular moment. So like the ball transfers its linear momentum to the bar, giving it angular momentum?
People on quora Say they are 2 different things, with different units of mesurement. But here it looks more like a matter of perspective . Did You find a solution? Im trying to figure this out myself.
does the ball has any angular velocity or it only has an angular momentum
By virtue of having angular momentum it must have an instantaneous angular velocity because Angular Momentum = Inertia times Angular Velocity. If Angular velocity was zero then there would be zero angular momentum. However, because we know the ball actually travels in a straight line, that angular velocity is usually expressed in linear terms as tangential (linear) velocity/radius
what is 1/3?
Moment of inertia of a rod about it's end is 1/3 ML^2
@@gautamgarg1578 How is that derived, though? There were no videos on that.
@@jackreid5970
The moment of inertia is a way to treat "a collection of point masses that rotate together" as a single object.
A rod is an infinite number of infinitesimally small point masses arranged in a line; we can calculate the angular momentum of the rod by adding each point up. To do this we'll need calculus. Let's use a 1-dimensional rod for simplicity.
Let the left end of the rod be at x=0, and the right end be at x=L.
A point mass located at x along the rod has a mass of dm. If we assume the rod has a uniform density λ (aside: λ = M/L), then the point mass dm is λdx (where dx is the length of an infinitesimally thin piece of the line).
The angular momentum of a single point is (dm)vx, where x is the distance from the axis at the left end of the rod.
If a rod is rotating at an angular velocity ω, a point at a distance x along the rod is moving at a speed v = ωx.
Let's add up the angular momentums of all the points: a continuous sum from x=0 to x=L of (λdx)(ωx)(x).
This is an integral, ∫λωx²dx from 0 to L evaluates to (λω)L³/3.
Plugging in λ = M/L, we get that the angular momentum of a 1d rod about its left end is (ML²/3)ω.
This quantity (ML²/3) would also arise if we talked about torque or other rotation stuff, so let's give it a name "moment of inertia" and it is special to this shape and size of rotating object. This is where I = (ML²/3), and L = Iω comes from.
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In general, I = ∫r²dm, where dm is a point mass in the object, and r is the distance of that point mass from the axis of rotation. To calculate this integral, you'll generally have to write dm in terms of an infinitesimal piece of space, like λdx or ρdV.
Wouldn't it make more sense to just say that in order for the ball to completely stop, an impulse acts on the ball from the rod, and in return the ball exerts the same impulse on the rod by Newton's second law. The math would still be exactly the same (angular momentum = mvr), and it's a lot more intuitive than saying the ball is constantly in angular motion about different axes and radii.