# Conservation of angular momentum

• WonderKitten
In summary, a homogeneous disc with a mass of 1.78 kg and a radius of 0.547 m is at rest on a polished surface, held in place by an axis. A particle with a mass of 0.311 kg and a velocity of 103 m/s collides with the disc at a point perpendicular to its surface, and becomes stuck at the impact point. The angular velocity of the system after the collision is -17.8. The calculations for this result were based on the conservation of angular momentum, using the cross product formula for angular momentum of a particle.
WonderKitten
Hi, I have the following problem:
A homogeneous disc with M = 1.78 kg and R = 0.547 m is lying down at rest on a perfectly polished surface. The disc is kept in place by an axis O although it can turn freely around it.
A particle with m = 0.311 kg and v = 103 m/s, normal to the disc's surface at the point of contact.
After the collision, the particle stays stuck to the impact point O'.
What's the angular velocity of the system after the collision?

It might be a bit confusing as I had to translate the problem, sorry about that. I'm not entirely sure if the disc is rotating on it's x-axis or z-axis (y-axis?), and by the particle being 'normal to the disc's surface' I think it means it's perpendicular to the disc? Not sure what I should do with this information.

So, since the axis of rotation isn't on the center of mass:
I (disc) = MR²/2 + MR² = 3MR²/2
The distance from O' to O: (using pythagoras)
r = 0.774
I (particle) = mr²

Ip * vR = - (Id + Ip) * wf
solving for wf and substituting the values I get wf = -10.6.

This is wrong, apparently. Not sure why, I'm out of ideas. Could someone help me out? Thanks!

[Moderator's note: Moved from a technical forum and thus no template.]

Last edited by a moderator:

The disk is rotating about point O in the plane of the screen. The particle collides with the disk in a direction perpendicular to the rim, i.e. the velocity is directed towards the center of the disk. This allows you to write the angular momentum of the particle relative to point O as ##L=mvR## which I think is what you have done, but I am not sure.

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WonderKitten
When the particle hits the disc it will begin rotating in the clockwise direction, that's why it's negative.
As for the calculations:
Ip = mr² = 0.186
Id = 3MR²/2 = 0.799
Ip + Id = 0.985
Ip * vR / Ip + Id = 10.6

If I use mVR = (Ip+Id)wf (as suggested by someone)
I get 17.8... but shouldn't I use 'mr²' instead of 'm'?
Thanks

Just saw your edit and I think that's it then? Thanks for the explanation on why it's mvR, got it.

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WonderKitten said:
get 17.8... but shouldn't I use 'mr²' instead of 'm'?
That's the number I got. Use mr2 instead of m in what expression? Angular momentum conservation says $$L_{\mathrm{before}}=L_{\mathrm{after}}$$ where$$L_{\mathrm{before}}=mvR; ~~L_{\mathrm{after}}=(I_d+I_p)\omega~~ \text{with} ~~I_d=\frac{3}{2}MR^2~~\text{and } I_p=m(2R^2).$$Perhaps you missed the point that the angular momentum of a particle is, by definition, the cross product ##\vec L =\vec r \times \vec p##, See here for example. In the present case, ##r ~\sin\theta =\text{constant}=R.##

Last edited:
kuruman said:
Perhaps you missed the point that the angular momentum of a particle is, by definition, the cross product ##\vec L =\vec r \times \vec p##, See here for example. In the present case, ##r ~\sin\theta =\text{constant}=R.##
Yep, that was it. Thank you!

## 1. What is conservation of angular momentum?

Conservation of angular momentum is a fundamental principle in physics that states that the total angular momentum of a system remains constant over time, unless acted upon by an external torque. This means that the rotational motion of a system will not change unless a force is applied to it.

## 2. How is angular momentum conserved?

Angular momentum is conserved because of the law of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In the case of angular momentum, this means that the total rotational inertia of a system will not change unless an external torque is applied.

## 3. What is the formula for calculating angular momentum?

The formula for calculating angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. This formula shows that the angular momentum is directly proportional to the moment of inertia and the angular velocity.

## 4. What are some real-life examples of conservation of angular momentum?

One example of conservation of angular momentum is the spinning of a figure skater. As the skater pulls their arms in, their moment of inertia decreases, causing their angular velocity to increase in order to maintain a constant angular momentum. Another example is the orbit of planets around the sun, where the angular momentum of the planets remains constant as they move in elliptical orbits.

## 5. How does conservation of angular momentum relate to the conservation of energy?

Conservation of angular momentum is closely related to the conservation of energy, as both are fundamental principles in physics. The conservation of angular momentum can be derived from the conservation of energy, as any change in angular momentum must be accompanied by a corresponding change in energy. Additionally, the law of conservation of energy can be used to explain the conservation of angular momentum in systems where no external forces or torques are present.

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