# Pendulum in a freely moving box

• fayled
In summary: So we can equate these two equations and solve for ##q##: ## q = c + k q^2 /2 ##.Not very much really - I just know the angles are small and so I can expand things using that.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
fayled

## Homework Statement

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

L=T-V
The EL equation

## 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.

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

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:
BvU said:
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 ?

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 ##

fayled said:
However I can't justify removing it, as small θ doesn't necessarily mean small (dθ/dt) does it?

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

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

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?

fayled said:
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?

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

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

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

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

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

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?

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

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

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

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 ##.

## 1. How does a pendulum behave in a freely moving box?

A pendulum in a freely moving box will continue to swing back and forth, with the amplitude and period of the swing depending on the length and weight of the pendulum. The motion of the box will not affect the motion of the pendulum as long as the box does not hit the pendulum.

## 2. What causes the motion of a pendulum in a freely moving box?

The motion of a pendulum in a freely moving box is caused by the force of gravity acting on the pendulum bob, which creates a restoring force that pulls the pendulum back towards the center.

## 3. Will the pendulum eventually stop swinging in a freely moving box?

Yes, the pendulum will eventually come to a stop due to the effects of friction and air resistance. The length and weight of the pendulum will affect how long it takes for the pendulum to stop swinging.

## 4. How does the motion of the box affect the motion of the pendulum?

If the box is moving with a constant velocity, the motion of the box will not affect the motion of the pendulum. However, if the box is accelerating or decelerating, this can affect the motion of the pendulum and may cause it to swing with a slightly different period or amplitude.

## 5. Can the pendulum's motion be used to measure the motion of the box?

In theory, yes, the motion of the pendulum could be used to measure the motion of the box. However, this would require precise measurements and calculations to account for the effects of the pendulum's own motion and the motion of the box on the pendulum's swing.

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