# Pendulum in a freely moving box

1. Mar 6, 2014

### fayled

1. The problem statement, all variables and given/known data
A box of mass M can slide horizontally on a frictionless surface. A simple pendulum of string length l and mass m, is suspended inside the block. Denote the coordinate of the centre of mass of the box by x and the angle that the pendulum makes with the vertical by θ . At t = 0 the pendulum displacement is θ = θ0 which is not equal to zero.

Find the Lagrangian and the equation of motion for the generalized coordinates x and θ . Which conservation law is obtained as a result of the cyclic coordinate?

Find the solutions for x and θ in the small angle approximation, hence show that the pendulum and the box execute SHO about their centre of mass at a frequency ω=[(M+m/M)^0.5]*(g/l)^0.5

2. Relevant equations
L=T-V
The EL equation

3. The attempt at a solution
The kinetic energy T=0.5M(dx/dt)2+0.5m[(dx/dt)2+l2(dθdt)2+2l(dx/dt)(dθ/dt)cosθ].

The potential energy measured from the support point of the pendulum U=-mglcosθ.

The Lagrangian L=T-V
L=0.5M(dx/dt)2+0.5m[(dx/dt)2+l2(dθdt)2+2l(dx/dt)(dθ/dt)cosθ]+mglcosθ

This gives the EOM as
(M+m)(d2x/dt2)+ml(d2θ/dt2)cosθ-ml(dθ/dt)2sinθ=0
(also x is a cyclic coordinate, which before taking the time derivative of (∂x/∂t), shows linear momentum is conserved), and
l(d2θ/dt2)+(d2x/dt2)cosθ=-gsinθ

In the small angle approximations, sinθ->θ, cosθ->1, then the EOM become
(M+m)(d2x/dt2)+ml(d2θ/dt2)-ml(dθ/dt)2θ=0
and
l(d2θ/dt2)+(d2x/dt2)=-glθ.

Then rearranging the second for (d2x/dt2)=-(gθ+l(d2θ/dt2)) and putting it into the first gives
d2θ/dt2=(gθ/l)[(m/M)+1-{ml(dθ/dt)2/Mg}]
i.e there is a nasty (dθ/dt)2 term which if not present would give me the correct answer... However I can't justify removing it, as small θ doesn't necessarily mean small (dθ/dt) does it?

Any clues? Thanks.

2. Mar 6, 2014

### BvU

What happens to the kinetic energy of $l \dot \theta \sin \theta$ ?
 Never mind, you have it in.

3. Mar 6, 2014

### BvU

At the risk of stumbling again: ${\partial T\over \partial \dot x} = (M+m)\dot x + m \ l \dot \theta \cos\theta\quad$ so why is the first EOM so complicated ?

Last edited: Mar 6, 2014
4. Mar 6, 2014

### fayled

Because I then need to use d/dt on that - which involves differentiating the following term by the product rule $m \ l \dot \theta \cos\theta\quad$

5. Mar 6, 2014

### voko

It does. In the small oscillations approximation, equations should be linear, anything non-linear is Taylor expanded, and only the linear terms are retained.

6. Mar 6, 2014

### fayled

I'm not quite sure what you mean. Do you mean I am allowed to ignore it just on the grounds it is non-linear?

7. Mar 6, 2014

### voko

Not because it is not linear. Because its Taylor expansion does not have any linear terms.

8. Mar 6, 2014

### fayled

Right, I'm not too sure how to go about Taylor expanding (dθ/dt)2 unfortunately - I should understand then.

9. Mar 6, 2014

### voko

It is already its own expansion. That is why it gets ignored.

10. Mar 6, 2014

### fayled

So I can ignore (dθ/dt)2 because it is not a linear term. Why can I ignore non-linear terms? Surely I should care about whether the quantity is small or not?

11. Mar 6, 2014

### voko

Has the method of the small oscillations approximation been explained to you? What is the theory behind it?

12. Mar 6, 2014

### fayled

Not very much really - I just know the angles are small and so I can expand things using that.

13. Mar 6, 2014

### voko

The Lagrangian is $L = T(q, \dot q) - U(q)$. It is further known that in all practical cases $T(q, \dot q) = t(q)\dot q^2$. Assuming $q = 0$ in equilibrium, we must have $U'(0) = 0$. So in small oscillations $U(q) \approx U(0) + U''(0)q^2/2 = c + k q^2 /2$. $T(q, \dot q)$ is approximated with $t(0)\dot q^2 = m \dot q^2/2$. The corresponding equation of motion is then $m \ddot q + k q = 0$.