# Quick Lagrangian of a pendulum question

• Lucy Yeats
In summary, the conversation discusses the use of the E-L equation to calculate the period of oscillation of a simple pendulum, and the effect of vertical acceleration of the pendulum support on its period of oscillation. The generalized coordinate θ is used to represent the angle of the pendulum, and the period of oscillation is found to be T=2pi√(l/g). The equations for the pendulum's motion with the support's acceleration are set up, with the y coordinate being -lcosθ+0.5at^2. The only freedom of the system is represented by the angle θ.
Lucy Yeats

## Homework Statement

Use the E-L equation to calculate the period of oscillation of a simple pendulum
of length l and bob mass m in the small angle approximation.

Assume now that the pendulum support is accelerated in the vertical direction at a rate
a, ﬁnd the period of oscillation. For what value of a the pendulum does not
oscillate? Comment on this result.

## The Attempt at a Solution

I've got the first bit:
L=(m/2)(l^2)(dθ/dt)^2-mgl(1-cosθ)
E.O.M.: d2θ/dt2+(g/l)sinθ=0
d2θ/dt2+(g/l)θ=0 in the small angle approximation,

For the next bit, I just need help setting up the equations:
So the generalized coordinates are θ and a.
Are the following correct?:
x=lsinθ
y=-lcosθ+at
(taking the origin as the point from which the pendulum is swinging)

Last edited:
Lucy Yeats said:
I've got the first bit:
L=(m/2)(l^2)(dθ/dt)^2-mgl(1-cosθ)
E.O.M.: d2θ/dt2+(g/l)sinθ=0
d2θ/dt2+(g/l)θ=0 in the small angle approximation,
Careful here, $\omega$ is that angular velocity in radians per time unit. Your T is the time to travel one radian (of the oscillatory cycle) not time per cycle. You need to multiply by $2 \pi$.

Thanks for pointing that out, I'll correct that in the first post. :-)

Any help would be great. :D

Lucy Yeats said:

Yes.

So the generalized coordinates are θ and a.

No.
"a" is not a coordinate.
Actually only θ is a generalized coordinate since the y coordinate of the support is constrained to be y=0.5at^2.
The angle θ is the only freedom that the system has.

Great, I've got it now.

Thanks! :-)

## 1. What is the Quick Lagrangian of a pendulum?

The Quick Lagrangian of a pendulum is a mathematical formula that describes the dynamics of a pendulum system. It takes into account the mass of the pendulum, the length of the string, and the angle at which the pendulum is displaced from its equilibrium position.

## 2. How is the Quick Lagrangian of a pendulum different from other Lagrangian formulations?

The Quick Lagrangian takes into account the instantaneous acceleration of the pendulum, making it more accurate for describing the motion of a pendulum that is rapidly changing. Other Lagrangian formulations may only consider the pendulum's average acceleration.

## 3. Why is the Quick Lagrangian of a pendulum important?

The Quick Lagrangian is important because it allows scientists and engineers to accurately model and predict the behavior of pendulum systems. This is useful in a variety of fields, including physics, engineering, and robotics.

## 4. How is the Quick Lagrangian of a pendulum used in real-world applications?

The Quick Lagrangian is used in a variety of real-world applications, such as designing pendulum-based clocks, analyzing the stability of bridges and building structures, and creating control algorithms for robotic arms.

## 5. Can the Quick Lagrangian of a pendulum be applied to other systems?

Yes, the Quick Lagrangian can be applied to other systems beyond just pendulums, such as double pendulums, swinging doors, and spring-mass systems. It can also be extended to more complex systems by combining it with other mathematical concepts, such as Hamilton's equations.

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