Lagrangian of a spring mass system

[tanaygupta2000]

Homework Statement

A particle of mass 'm' is tied to one end of a massless spring (spring constant k and unstretched length r0). The other end of the spring is fixed to a point P on a smooth horizontal plane on which this particle is moving. If the instantaneous position of this particle is (r,θ), then obtain the Lagrangian and Hamiltonian of the ststem. Also find equations of motion of the system.

Relevant Equations

Lagrangian, L = KE - PE
Hamiltonian, H = Σpx(dot) - L
Euler Lagrange equation, ∂L/∂q - d/dt (∂L/∂q(dot)) = 0

I know that from the given problem, I need to find the expression for Kinetic energy,
KE = 1/2 m [r(dot)]^2

and Potential energy,
PE = 1/2 k r^2

So L = 1/2 m [r(dot)]^2 - 1/2 k r^2
Hence H = 1/2 m [r(dot)]^2 + 1/2 k r^2

I assume that the fixed length r0 is provided to find the value of end constants.
But what is the function of theta? Is it a generalized coordinate ?

 

[BvU]

You use r for both the vector and the scalar -- that is confusing

But what is the function of theta? Is it a generalized coordinate ?

Does it appear in the kinetic energy ?

 

[tanaygupta2000]

You use r for both the vector and the scalar -- that is confusing
Does it appear in the kinetic energy ?

Sir r only defines the instantaneous distance of the particle from point P.
I think KE depends only on velocity r' and not on theta or any of its derivatives.

 

[TSny]

I think KE depends only on velocity r' and not on theta or any of its derivatives.

One possible motion would be circular motion about P. How would you express the KE for this case?

 

[tanaygupta2000]

One possible motion would be circular motion about P. How would you express the KE for this case?

KE = (1/2)mr'² + (1/2)mθ'²

 

[TSny]

KE = (1/2)mr'² + (1/2)mθ'²

Close. Check the dimensions of the second term.

 

[tanaygupta2000]

Close. Check the dimensions of the second term.

Sorry sir
It is (1/2)mr'² + (1/2)mr0²θ'²

 

[tanaygupta2000]

I think the system can be considered as the particle executing a circular motion around point P as center in a horizontal plane, connected to the point P by a massless spring of unstretched length r_o and spring constant k. The instantaneous distance of the particle from P is r (greater than r_o due to stretching of spring) and instantaneous angle from axis is θ (due to rotation of spring as a radius around P).
I don't think Potential energy of the particle depends on θ or its derivative, it should be simply (1/2)kr²

 

[TSny]

Circular motion is only a special case. For the general case, both $r$ and $\theta$ will depend on time. You have the correct expression for KE if you use $r$ instead of $r_0$ in the second term of KE. The PE should contain both $r$ and $r_0$.