# Regular Elliptic Pendulum - Help Please

1. Mar 9, 2012

### urkirsty

1. The problem statement, all variables and given/known data

A uniform rod AB of mass M and length 2l attatched to a slider at A which is constrained to move along a smooth horizontal wire. The rod, which is subject to gravity, is free to swing in the vertical plane containing the wire. Using Ω (the angle between rod and vertical) and x coordinate of the centre of mass, find an expression for the kinetic and potential energies, deduce Lagrange's equations. Show that the Lagrangian is regular.

3. The attempt at a solution

I've been trying to get my head around this for 2 days but I've never seen anything like it before so I just don't know what to do.

I know I have to treat the rod as a continuum, but I've never done this before so I'm not sure how to. Also the coordinate system is one I've never seen before.

Sorry it doesn't seem like I've done much but I've tried every way I know and nothing's working, any advice would be greatly appreciated. Thank you

2. Mar 9, 2012

### tiny-tim

welcome to pf!

hi urkirsty! welcome to pf!

try this part first …

… what do you get?

3. Mar 9, 2012

### urkirsty

Well I have an equation for the Kinetic energy now, I've taken the kinetic energy in relation to the centre of mass and then added the kinetic energy from the rotation with the moment of inertia.
I have (x. means x differentiated wrt t, I wasn't sure how to show it on here)

KE = 1/2[Mx.^2 + θ.^2 Ml^2 sin^2(θ) + 1/3 θ.^2 Ml^2]

and Potential energy V = -Mglcosθ (taking up to be +ve)

Sorry those equation look really messy typed up here. I'm hoping they're at least on the right lines, fingers crossed!

4. Mar 9, 2012

### tiny-tim

hi urkirsty!

(use x' for derivative, and try the X2 button just above the Reply box )

yes, that's fine

(except you need 2l at the end, not l)

now go for lagrange's equations!​

5. Mar 10, 2012

### imjess

hi...what was your x and y?

was it y=-2lcosθ and x? and isnt your potential energy suppose to be v = -mg2lcosθ

6. Mar 12, 2012

### imjess

KE = 1/2[Mx.^2 + θ.^2 Ml^2 sin^2(θ) + 1/3 θ.^2 Ml^2]

How did you get the last part of KE.....no can only get KE = 1/2[Mx.^2 + θ.^2 Ml^2 sin^2(θ)]

7. Mar 13, 2012

### Nathen

I have the same problem as above, not knowing where that 1/3 θ.^2 Ml^2 term comes from, can anyone please help?

8. Mar 13, 2012

### tiny-tim

welcome to pf!

hi imjess! hi Nathen! welcome to pf!
that's the rotational kinetic energy

KEtotal = KEtranslational + KErotational

= 1/2 mvc.o.m2 + 1/2 Ic.o.mω2

where Ic.o.m is the moment of inertia about the centre of mass, and ω is the angular velocity

(in this case, Ic.o.m = (1/12)M(2l)2)

we can also write KEtotal = 1/2 Ic.o.rω2 where Ic.o.r is the moment of inertia about the centre of rotation

but in this case, we don't know where the centre of rotation is!

9. Mar 13, 2012

### imjess

Thank you so much!!

And is v= -2lmgcosθ. Just to check if that's right...

10. Mar 13, 2012

### tiny-tim

no, PE = -lmgcosθ

(you use the height of the centre of mass of the uniform rod of length 2l)