Degrees of freedom and holonomic constraints

[Lambda96]

Homework Statement

See Picture

Relevant Equations

none

Hi,

I am not quite sure if I have solved task 2a and 2b correctly.

For task 2a I would say, because of the constraints, that the system has only 1 degree of freedom. Since the vectors must always have an angle of Pi/4 to each other, this would mean that if one vector moves up, the other must also move up. If one of the vectors moves to the right, the other must also move to the right, and so on.

For 2b, I would say that since the vectors are moving on the surface of the sphere, one vector can move in two directions at a time, independent of the movement of the other vector, so 2 degrees of freedom with 2 vectors gives a total of 4 degrees of freedom

Have I solved both tasks correctly or have I made a mistake somewhere?

 

[BvU]

Hi,

For task 2a I would say

Isn't task 2a saying "without any constraints" ? I.e. without any constraints at all ? Yet you find only one degree of freedom ?

For task 2b I would subtract the number of degree of freedom in the constrained system from the answer as found in 2a...

I suggest/propose you first do the exercise with only one particle that is constrained to move on the surface of the sphere. After that, what changes when a second particle is added with the extra one $\pi\over 4$ constraint ?

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[Lambda96]

Thanks BvU for your help

I have read through the task again more carefully and have completely misunderstood task part a and b, sorry for that.

Task a)
1 particle has a total of 6 degrees of freedom, 3 with respect to translation and 3 with respect to rotation. Since I have two particles, there are a total of 12 degrees of freedom

Task b)
There are two constraints in total. The first that the particles must move on the surface of the sphere and the second that the angle between them must always be pi/4.

Task c)
1 is the constraint for the surface on the sphere and 2 for the angle between the two vectors

Task d)
With the condition that a particle moves on the surface, of the sphere, you can see the equation from 1c, that now the z-coordiante depends on x and y and so 1 degree of freedom is lost and a particle has only 5, so two particles have 10 degrees of freedom.
Concerning the angle, I am not quite sure. Before the second constraint, the two particles could move independently, now they are connected. Through which one of the vectors is always dependent on the other and thus has only 1 degree of freedom. In total, the system has 6 degrees of freedom.

 

[PhDeezNutz]

I think for a “particle” (which I read as a “point particle”) you don’t consider rotational degrees of freedom because they don’t have any physical extent. So I get 6 degrees of freedom for part A.

 

[Steve4Physics]

Task a)
1 particle has a total of 6 degrees of freedom, 3 with respect to translation and 3 with respect to rotation. Since I have two particles, there are a total of 12 degrees of freedom

An object of finite size has 6 degrees of freedom. E.g. you can spcifiy the exact configuration of an aeroplane by 6 values: x, y and z cooordinates of it's centre of gravity and 3 rotation angles about the x, y, and z axes (roll, pitch yaw).

But a 'particle' has no size - it is a point. 'Rotation' of a particle is meaningless - there is no way to disinguish a particle before and after a rotation about its 'centre'.

Task b)
There are two constraints in total. The first that the particles must move on the surface of the sphere and the second that the angle between them must always be pi/4.

Particles are treated as independent entities. Each particle must be on the sphere's surface. That's two constraints to start with!

Task c)
1 is the constraint for the surface on the sphere and 2 for the angle between the two vectors
View attachment 316774

I think the left side of your equation says 1/√2, which is dimensionless.
But the right side has dimensions of length.
So the equation is inhomogeneous and must be wrong.
And you haven't shown how you derived it.
Edit. Many apologies. I misread the blob between the square root terms as a plus sign

Task d)

With the condition that a particle moves on the surface, of the sphere, you can see the equation from 1c, that now the z-coordiante depends on x and y and so 1 degree of freedom is lost and a particle has only 5, so two particles have 10 degrees of freedom.
Concerning the angle, I am not quite sure. Before the second constraint, the two particles could move independently, now they are connected. Through which one of the vectors is always dependent on the other and thus has only 1 degree of freedom. In total, the system has 6 degrees of freedom.

You are meant to answer this using your answers from parts a) and b).
(DoF for constrained system) = (DoF for unconstrained system) - (number of constraints)
(That gives you the number of generalised coordinates you need. for part e).)

 

[BvU]

read through the task again more carefully

Always a good idea

There are two constraints in total. The first that the particles must move on the surface of the sphere and the second that the angle between them must always be pi/4

That way my approach for 2b doesn't work. Personally I would count three constraints for the coordinates: one for each particle plus one for the angle.

[ah, Steve is helping as well].

 

[haruspex]

An object of finite size has 6 degrees of freedom.

There are also the cases
- thin rod (e.g. diatomic molecule, no internal vibration), 5 degrees
- lamina (e.g. triatomic molecule, ditto), 6 degrees

 

[haruspex]

the right side has dimensions of length.

Looks dimensionless to me. L² top and bottom.

 

[Steve4Physics]

Looks dimensionless to me. L² top and bottom.

Aargh. You're right. I misread the blob between the 2 square roots in the denominator as a plus sign. I'll amend my post.

 

[BvU]

So, @Lambda96, I take it you have sorted out parts a and b.
For part c, I think the exercise composer wants you to answer in terms of the coordinates provided. He/she even explicitly says so. So not $x^2+y^2+z^2=R^2$ but $x_1^2+y_1^2+z_1^2=R^2$ and $x_2^2+y_2^2+z_2^2=R^2$ for the first two and your encircled number 2 for the third. Well done. On to part d.

Concerning the angle, I am not quite sure. Before the second constraint, the two particles could move independently, now they are connected. Through which one of the vectors is always dependent on the other and thus has only 1 degree of freedom. In total, the system has 6 degrees of freedom.

A way to look at this: with the first constraint the position of particles 1 on the sphere can be pinpointed with two generalized coordinates. Example: latitude and longitude on earth. Exexcute an imagined coordinate transform that makes that first particle occupy the north pole. The second particle is now constrained to the new equator, and a generalized coordinate for the second particle is then its new longitude. (No need to actually do any calculations, just counting is enough).

Inadvertently this example makes part e a breeze indeed !

A nice exercise, this is .

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[Lambda96]

Thank you all for your help.

I have one more question about the constraints. Regarding that the two particles must move on the surface, can I combine these constraints into one or do they have to be specified separately?

 

[BvU]

They are completely separate.

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[Lambda96]

Thank you BvU

So I have a total of 6 degrees of freedom, 3 constraints whereby the degrees of freedom can be calculated as follows $$3N-k=f$$ $$with$$ $$N=2, k=3 $$ $$get$$ $$3*2-3=3$$ 3 degrees of freedom

Is that right?

 

[BvU]

You tell us !

If there is something unclear, ask. "Is that right" doesn't feel good in the PF context.
The idea is that you get help until you can convince yourself that your result is the right one.

I find my example quite convincing. Best I can do. But I've made errors before.

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[Steve4Physics]

Exexcute an imagined coordinate transform that makes that first particle occupy the north pole. The second particle is now constrained to the new equator, ...

It's worth noting that the angle between the particles’ position vectors (with the origin at the centre of the sphere) is $\frac {\pi}4$ and not $\frac {\pi}2$.

With a coordinate transformation which shifts one particle to a new north pole, the other particle won’t be constrained to the new equator - it will be constrained to the new 45ºN circle of latitude.

 

[BvU]

Oops, what a blunder ! Thanks, steve!