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Proving rotational motion using newton's laws

by metalrose
Tags: laws, motion, newton, proving, rotational
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metalrose
#1
Aug8-10, 12:43 AM
P: 126
I have asked this question before, but couldn't get a satisfactory response.
Let me make the problem more concise.

We have a two particle system, i.e. two masses of mass m each joined together by a spring of spring constant k.
A force F is applied to one of these particles in a direction perpendicular to that of the spring joining the two masses.
As the spring constant tends to infinity, the system behaves as two masses joined by a rigid rod.

Prove, (without using the kinematic approches of torque, ang. momentum et al) that the trajectory followed by the two masses, as viewed from the C.O.M. frame would be a circle centered at the C.O.M. and find out the angular velocity or acc.

You can not use the usual approches of finding the torque.
Prove the trajectory using anything but the torque or ang. momentum eq.'s

Even if you dont solve this here, please atleast tell me wether it is solvable or not. And if it is, what topics are needed as a prerequisite?
Can lagrangian mechanics solve my above problem? or theories which deal with elasticity?
And if they can, please give me some references so that I can look up.

Thanks
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Filip Larsen
#2
Aug8-10, 06:24 AM
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You should be able to sketch a "proof" as seen from the CM using the fact that the initial velocity of the masses are perpendicular to the rod which means that the velocity vector will only turn (and not increase or decrease in magnitude) and the speed is thus conserved. I guess will need to identify some invariant related to the geometric relationship between the two masses (the distance constraint imposed by the rod) in order conclude the rod forces are always perpendicular to the velocity of the masses and that they therefore are bound to a constant rotation around the CM.
metalrose
#3
Aug8-10, 07:18 AM
P: 126
@Filip larsen

Yes, the initial velocity vector of both the particles is perpendicular to the rod, "but in the same direction". And even if the rigid rod were to provide them with a centripetal force so as to rotate them, since they have an initial velocity in the same direction and centripetal forces in the opposite directions, one particle would have a tendency to rotate clockwise, while the other, anticlockwise.

And this ofcourse can't be true, because in the rotational motion we observe, both particles eitehr move clockwise or anticlockwise.
There lies the problem.
Hope you can help.

Thanks

Filip Larsen
#4
Aug8-10, 07:31 AM
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Proving rotational motion using newton's laws

Quote Quote by metalrose View Post
Yes, the initial velocity vector of both the particles is perpendicular to the rod, "but in the same direction".
No, as seen from the CM the two masses have opposite velocities.

To see this you can look at it from a frame that is at rest with the masses just before the impulsive force is applied. Right after the impulsive force has been applied the first mass moves perpendicular to the rod but the second mass is at rest since none of the impulsive force can be translated to the second mass. If the rod was missing then the first mass would continue to move in its initial direction and the second mass would continue to stay at rest. The CM of the two masses (without the rod) would therefore travel with half the velocity of the first mass, which means, that seen from the CM the first mass moves one way while the other moves the opposite way. Now "insert" the rod again and note that at the CM the rod forces cancels each other, so the CM will continue to move with this initial "half" velocity even with the rod present.

From here you should be able to continue the argument with perpendicular forces on the masses.
metalrose
#5
Aug8-10, 08:53 AM
P: 126
@filip larsen

I guess this won't be the case.
If we have a rigid rod, the force applied on any one of the particles will get evenly distributed all throughout the rod and upto the other particle and this is how the entire system will achive a common acc. in the forward direction given by a=F/2m. So the initial acc. of both the particles is the same, and so both should have the same initial velocity.

Did I go wrong somewhere?
arildno
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Aug8-10, 09:34 AM
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Quote Quote by metalrose View Post
I have asked this question before, but couldn't get a satisfactory response.
You have gotten satisfactory response to this.
Period.
Filip Larsen
#7
Aug8-10, 09:35 AM
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Quote Quote by metalrose View Post
If we have a rigid rod, the force applied on any one of the particles will get evenly distributed all throughout the rod and upto the other particle and this is how the entire system will achive a common acc. in the forward direction given by a=F/2m. So the initial acc. of both the particles is the same, and so both should have the same initial velocity.
This is not so. The (idealized one-dimensional) rod is only capable of transferring forces along the direction of the rod so there is no way a perpendicular impulsive force on one mass can yield a perpendicular force on the other mass at the same time. You can replace the rod with a thin strong thread that is incapable of transferring anything but "pull-forces" and still get the same motional behavior as in the original setup.


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