Angular momentum and Linear momentum elastic collision

In summary: If there's no torque, why would ω change?PS: This is unrelated but could the admins please remove the automatic adding of the template each time the thread is previewed, its really annoying :) (or maybe its just a problem with vBulletin)
  • #1
Unperfect
4
0

Homework Statement


Ok so I have two objects of different masses [itex]{m}_{{1}}[/itex] for the projectile and [itex]{m}_{{2}}[/itex] for the spinning top. The projectile has only translational kinetic energy in the direction of the arrow (see attached image), with [itex]{v}_{{1}}[/itex] and [itex]{v}_{{2}}[/itex] as initial and final linear velocities, respectively the initial velocity is known but not the final one. It collides with a spinning top, which has [itex]{\omega }_{{1}}[/itex] and [itex]{\omega }_{{2}}[/itex] as initial and final angular velocities, the initial being known.
.

Homework Equations


Firstly there is no torque(I have already asked a few people and they mentioned torque), as the perpendicular distance of the projectile from the axis of the spinning top is 0.(so [itex]\tau =r\times F = 0[/itex])

Energy wise: [itex]{E}_{{before}}=\frac {1}{2}I{\omega }_{{1}}^{{2}}+\frac {1}{2}{m}_{{1}}{v}_{{1}}^{{2}}[/itex]

[itex]{E}_{{after}}=\frac {1}{2}{m}_{{1}}{v}_{{2}}^{{2}}+\frac {1}{2}{m}_{{2}}{v}_{{s2}}^{{2}}+\frac {1}{2}I{\omega }_{{2}}^{{2}}[/itex]
vs2 is the final velocity of the spinning top.
The http://en.wikipedia.org/wiki/Elastic_collision" [Broken] are also relevant.


The Attempt at a Solution


Well I know how to solve collisions, but only when they are exclusively linear or angular.

I have basically used the conservation of energy principle, solving first in 1D(is that possible, or do I have to solve in 2d? And in 2d, how would you do it taking account of the angular velocity of the spinning top?) which obviously left me without answer.

Considering solving in 2D, I based myself on this:
http://www.myphysicslab.com/collision.html
but I am not sure it would work for a spinning top, could anyone advise me on how to use these equations but with a spinning top?
So basically I am a lost, could anyone help out/ give me some possible directions to explore?

Thanks in advance...
PS: This is unrelated but could the admins please remove the automatic adding of the template each time the thread is previewed, its really annoying :) (or maybe its just a problem with vBulletin)
 

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  • #2
Welcome to PF!

Hi Unperfect! Welcome to PF! :smile:

(have an omega: ω :wink:)
Unperfect said:
Firstly there is no torque(I have already asked a few people and they mentioned torque), as the perpendicular distance of the projectile from the axis of the spinning top is 0.(so [itex]\tau =r\times F = 0[/itex])

If there's no torque, why would ω change? :confused:
PS: This is unrelated but could the admins please remove the automatic adding of the template each time the thread is previewed, its really annoying :) (or maybe its just a problem with vBulletin)

Do you mean that if you preview n times, you get n+1 templates?
 
  • #3


tiny-tim said:
Hi Unperfect! Welcome to PF!

(have an omega: ω:wink:)
Thanks!(and thanks for the ω also, I couldn't be bothered to find the ascii code for it :P)
tiny-tim said:
If there's no torque, why would ω change? :confused:
Thats maybe were I'm getting things wrong, but maybe I should explain a bit further; I am also doing this as an experiment for a project and in the experiment the spinning top's ω reduces upon impact(by about 200-300 rpm my tachometer tells me, with a reasonably fast projectile) and "ejects" the projectile back at an angle(around pi/4 rad). But having the data is not enough(well for me at least, maybe the teacher just wants that), I also want to know the theory behind(and possibly program a simulation).
So anyway while googling and asking away about change in ω, I found that most people's reply involved torque but for torque to be non-zero(hence have an impact and be useable) the force applied by the projectile has to be at a distance perpendicularly from the spinning top's axis. But ideally(i'm sure its not perfect in my experiment) this distance is zero(see diagram in first post).
Maybe I should ask the question like this:
If I were to program a simulation of a collision in 2D of two objects of known mass and initial velocity and ω, one with exclusively KE and the other with exclusively RKE, what equations should I use to calculate the outcome?(most video games easily manage that but I don't know how :P) Does friction have to be taken into account?
Sorry if I'm making complicated but help is very much appreciated :approve:
Do you mean that if you preview n times, you get n+1 templates?
Thats exactly it, though I think it does check to see if the template is there(character by character) and if it's not(or has changed) it appends a template.
 
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  • #4
Unperfect said:
I am also doing this as an experiment for a project and in the experiment the spinning top's ω reduces upon impact(by about 200-300 rpm my tachometer tells me, with a reasonably fast projectile) and "ejects" the projectile back at an angle(around pi/4 rad).

If I were to program a simulation of a collision in 2D of two objects of known mass and initial velocity and ω, one with exclusively KE and the other with exclusively RKE, what equations should I use to calculate the outcome?(most video games easily manage that but I don't know how :P) Does friction have to be taken into account?

Hi Unperfect! :smile:

(ooh, i didn't give you the full list :rolleyes: … have a pi: π :wink:)

If the top "ejects" the projectile back at about 45º, then it looks as if friction is involved.

And perhaps you should try to measure the coefficient of restitution (the inelasticity) between the top and the object, with the top fixed (and rotating at various speeds, including zero)?

(btw, are you doing experimental physics, or computer science? In other words, are you actually trying to find out what happens, or are you just practising computer modelling?)
Thats exactly it, though I think it does check to see if the template is there(character by character) and if it's not(or has changed) it appends a template.

I've noticed that some members post two or even three templates, and I've always wondered why! :rolleyes:

After a quick search in the Feedback sub-forum, I now see it's a well-known bug that we just have to put up with … see https://www.physicsforums.com/showthread.php?t=216095&highlight=template+homework :wink:
 
  • #5
tiny-tim said:
(ooh, i didn't give you the full list … have a pi: π )
Thanks:smile:, but I do use my wacom tablet which(quite accurately surprisingly) transcripts whatever math I write in it into latex code so I can my symbols if I really need them :P.
tiny-tim said:
If the top "ejects" the projectile back at about 45º, then it looks as if friction is involved.
Yeah, friction is involved, it can't have possibly been something else; what was i thinking?
Anyway I decided to try a second approach, I'm not sure if it's a correct reasoning though.

Can I consider a single point on the spinning top(the one where the impact with the projectile occurs.[ie, red point on attached diagram]) and consider it to have a linear velocity(as demonstrated by the green line) calculated using v=rω.
Then can I consider that point on the spinning top to have a linear momentum of [itex]
{p}_{{2}}={m}_{{2}}r{\omega }_{{1}}
[/itex]?
With this I can resolve in 2D the collision that occurs and sure enough if I do this, torque is present(well a direction in that movement at least).
I solve a sample problem assuming 100% elasticity and I got a similar result to what you can see in diagram 3(the arrows with a black stroke are the resultant velocities).

No I have another problem; I do not know the magnitude of the force to resolve
[itex]
\tau =r\times F
[/itex].
because I have no measure of the time of impact, I can only calculate the change in momentum...

tiny-tim said:
And perhaps you should try to measure the coefficient of restitution (the inelasticity) between the top and the object, with the top fixed (and rotating at various speeds, including zero)?

(btw, are you doing experimental physics, or computer science? In other words, are you actually trying to find out what happens, or are you just practising computer modelling?)
I'm not sure exactly if I can measure the restitution coefficient, I do have two photogates so I could measure it but that might complicate the process, I might assume it to be 1 and then comment on that assumption in the evaluation.

And I'm doing experimental physics, but I would like to program a simulation so as to compare the results to comment(this is for an Extended Essay[a component of the IB Diploma]).

Thanks
 

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  • #6
Hi Unperfect! :smile:

I'm very sorry I haven't replied earlier …

your pictures hadn't come up yet when I first looked, nor when I checked back, and after that I forgot to check again. :redface:
Unperfect said:
Can I consider a single point on the spinning top(the one where the impact with the projectile occurs.[ie, red point on attached diagram]) and consider it to have a linear velocity(as demonstrated by the green line) calculated using v=rω.
Then can I consider that point on the spinning top to have a linear momentum of [itex]
{p}_{{2}}={m}_{{2}}r{\omega }_{{1}}
[/itex]?
With this I can resolve in 2D the collision that occurs and sure enough if I do this, torque is present(well a direction in that movement at least).
I solve a sample problem assuming 100% elasticity and I got a similar result to what you can see in diagram 3(the arrows with a black stroke are the resultant velocities).

I really don't like short-cuts like that …

they probably don't work (and does it take into account the rotational energy of both bodies?), but even if they do, it tends to take longer to check that they work than it would to do it the standard way.
Now I have another problem; I do not know the magnitude of the force to resolve
[itex]
\tau =r\times F
[/itex].
because I have no measure of the time of impact, I can only calculate the change in momentum...

I'm not sure exactly if I can measure the restitution coefficient, I do have two photogates so I could measure it but that might complicate the process, I might assume it to be 1 and then comment on that assumption in the evaluation.

You can calculate the torque (or rather the torque-impulse) from the change in energy of the top.

(But do you need the torque? Isn't conservation of angular and linear momentum enough?)

I think you do need a realistic figure for the restitution coefficient … say to the nearest .05 … so that you can write a reasonably accurate conservation of energy equation (and don't forget that the friction will rotate the projectile also. :wink:).

I suspect that finding the restitution coefficient with a non-spinning top is probably accurate enough. :smile:
 
  • #7
tiny-tim said:
Hi Unperfect! :smile:
I think you do need a realistic figure for the restitution coefficient … say to the nearest .05 … so that you can write a reasonably accurate conservation of energy equation (and don't forget that the friction will rotate the projectile also. :wink:).

I suspect that finding the restitution coefficient with a non-spinning top is probably accurate enough. :smile:
Thank you for your help; I can get the restitution coefficient quite easily actually.
But again the problem I have is calculating this collision theoretically; but I'll sort that out with my physics teacher I think.
If you think you can help, well all I need is a way to calculate the outcome of a collision of objects with rotational and kinetic energy(that is, ignoring all the post collision data I can collect) in 2 dimensions, provided I know all the starting variables(ie. elasticity coefficient, celocity, angular velocity,mass etc.) because I have yet to find a physics textbook which provides me with equations/guidance on handling such problems.

Anyway, thank you very much for your help.
 

1. What is the difference between angular momentum and linear momentum in an elastic collision?

Angular momentum is a measure of rotational motion, while linear momentum is a measure of translational motion. In an elastic collision, both angular and linear momentum are conserved, meaning that the total amount of each type of momentum before the collision is equal to the total amount after the collision.

2. How does angular momentum affect the direction of an object in an elastic collision?

In an elastic collision, the direction of an object's angular momentum will change depending on the direction of its linear momentum and the location of the collision. If the linear momentum is perpendicular to the collision location, the angular momentum will change in the same direction. If the linear momentum is parallel to the collision location, the angular momentum will change in the opposite direction.

3. What factors can affect the amount of angular momentum in an elastic collision?

The amount of angular momentum in an elastic collision can be affected by the mass, velocity, and location of the objects involved in the collision. The moment of inertia, which is a measure of an object's resistance to rotational motion, also plays a role in determining the amount of angular momentum.

4. Can the total angular momentum be conserved in an inelastic collision?

No, in an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound. This means that the total angular momentum will not be conserved, as some energy is lost in the collision.

5. How is the conservation of angular momentum related to the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted into different forms. In an elastic collision, the conservation of angular momentum is related to the conservation of energy because the total amount of energy before and after the collision must be the same, even though it may be in different forms.

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