Degrees of freedom with a particle and a rod

[Kaguro]

Homework Statement

Given that a mass-less rod is free to move in space. And a particle is constrained to move on it. How many degrees of freedom are there?

Relevant Equations

None

The rod itself should have 3 translational+2 rotational DOF.
The particle on top of the rod has one additional DOF.

So total should be 6. But answer given is 4.

What I'm thinking wrong?

 

[PeroK]

Well, I guess it doesn't actually say that it's free to rotate?

 

[Kaguro]

Well, I guess it doesn't actually say that it's free to rotate?

No, the question just said free to move in space.
I assumed movement involves rotation too, shouldn't I?

 

[PeroK]

No, the question just said free to move in space.
I assumed movement involves rotation too, shouldn't I?

How else do you get 4 degrees of freedom? I think the question is ambiguous.

 

[Kaguro]

How else do you get 4 degrees of freedom? I think the question is ambiguous.

Yes, it is.

Okay, thanks for helping!

 

[Delta2]

if one end of the rod is kept fixed, then the answer is indeed 3+1=4 because we just have 3 DOFs for the free end of the rod and 1 for the particle constrained on rod. However if both ends are free to move then i believe the answer is 3+3+1=7.

 

[PeroK]

if one end of the rod is kept fixed, then the answer is indeed 3+1=4 because we just have 3 DOFs for the free end of the rod and 1 for the particle constrained on rod. However if both ends are free to move then i believe the answer is 3+3+1=7.

This is not right. There are only two rotational degrees of freedom for a rod. E.g. polar and azimuthal angles.

 

[Kaguro]

This is not right. There are only two rotational degrees of freedom for a rod. E.g. polar and azimuthal angles.

That's what I was thinking.

The rotation along the axis of rod shouldn't change anything.

 

[Delta2]

Ok i was wrong, if both ends are free then we have 6 DOFs, but what about if one end is kept fixed, then we just have 4 DOFs, don't we?

 

[Kaguro]

if one end of the rod is kept fixed, then the answer is indeed 3+1=4 because we just have 3 DOFs for the free end of the rod and 1 for the particle constrained on rod. However if both ends are free to move then i believe the answer is 3+3+1=7.

What do you mean by fixing one end, though?

Shouldn't that stop the translation of rod?

 

[Delta2]

What do you mean by fixing one end, though?

Shouldn't that stop the translation of rod?

Well yes, that's correct.

 

[Delta2]

OK i see, i was wrong again, if one end is fixed then the answer is 2+1, if both ends are free then it is 5+1 as was said in the OP afterall.

 

[Delta2]

On second thought it could be 3+1 in the case of one end fixed, IF the length of the rod is variable (the problem statement doesn't say anything about that).

 

[Kaguro]

On second thought it could be 3+1 in the case of one end fixed, IF the length of the rod is variable (the problem statement doesn't say anything about that).

Please explain to me, how would variable length of rod add one more DOF?

 

[Delta2]

Please explain to me, how would variable length of rod add one more DOF?

Think of it like a rubber rod or something, a rod that is a spring on the same time.

 

[PeroK]

Think of it like a rubber rod or something, a rod that is a spring on the same time.

What is the purpose of this digression? In what way is it homework helping?

 

[Kaguro]

Think of it like a rubber rod or something, a rod that is a spring on the same time.

The rod is massless.

If I consider it as a spring, then... It should have 2 more DOF. Kinetic and potential.

Or maybe not?

 

[Delta2]

The rod is massless.

If I consider it as a spring, then... It should have 2 more DOF. Kinetic and potential.

Or maybe not?

I don't think there is something like a potential DOF. DOF is for kinetic degree of freedom, like in how many different ways it can move.
But anyway I think I 've put us on the wrong path, the problem statement probably implies that the rod is rigid as we all have a rigid thing in mind when we talk about a rod.

 

[Kaguro]

Okay!

Thanks everyone.

 

[haruspex]

It is not even clear whether this refers to positional degrees of freedom, kinetic degrees of freedom or both.
It is interesting we are told the rod is massless. Can't think how that is relevant.

 

[Kaguro]

It is not even clear whether this refers to positional degrees of freedom, kinetic degrees of freedom or both.
It is interesting we are told the rod is massless. Can't think how that is relevant.

Well, I found this question in the chapter on Lagrangian mechanics. So I think it refers to only positional DOF, the number of generalized coordinates needed to specify the configuration of the system.
(My bad for not giving out this context)

 

[wrobel]

3 degrees of of freedom. Because it is a particle in the space. At least it is the only way to make the statement of the problem correct.
Else you will get degenerate kinetic energy in the Lagrange equations

 

[haruspex]

3 degrees of of freedom. Because it a particle in space. At least it is only way to make the statement of the problem correct.
Else you will get degenerate kinetic energy in the Lagrange equations

Are you saying that because the rod is massless it is irrelevant to the degrees of freedom?

 

[wrobel]

Are you saying that because the rod is massless it is irrelevant to the degrees of freedom?

It is not exactly what I said.

 

[Kaguro]

Assuming the rod is parallel to x axis.
For now, particle is at center of the rod.

If the rod moves right by δx, and particle moves left by δx, does this constitute a new state?

 

[wrobel]

We do not have dynamical equations to describe rotation of the massless rod

 

[well_wisher]

Well, I found this question in the chapter on Lagrangian mechanics. So I think it refers to only positional DOF, the number of generalized coordinates needed to specify the configuration of the system.
(My bad for not giving out this context)

Can you write the name of book please?