Simple Pendulum in an elevator

[Jumblebee]

Homework Statement

Using the usual angle $$\phi$$, as generalized coordinate, write down the Lagrangian for a simple pendulum of length l suspended from the ceiling of an elevator that is accelerating upward with constant acceleration a. Find the Lagrangian equation of motion and sho that it is the same as that for a normal, nonaccelerating pendulum except that g has been replaced by g+a.

Homework Equations

Lagrangian is L=Kinetic energy- potential energy=T-U

The Attempt at a Solution

I need to find the kinetic energy and potential energy, which i know that T=1/2mv² and U=mgh.
So I think that T=1/2ml²$$\phi$$ ' ² (the phi has a prime, its hard to see) and
U=mgl(1-cos$$\phi$$)
However, I don't think that these are right because I didn't do anything with the fact that it is in an elevator accelerating. Any help?

 

[turin]

Your generalized coordinate system is inconsistent with the motion of the elevator. The potential energy will change for two reasons: 1) the pendulum bob swings up and down, and 2) the elevator changes elevation. The kinetic energy can "always" be written as T=(1/2)mv^2, but this v is no longer simply an angular motion because you also have to account for the motion due to the changing elevation of the elevator.

 

[Jumblebee]

Okay, so I need to fix my velocity, but I don't really know what to do with it, the only thing i know about it is that it is moving up with a constant acceleration, but then what happens with the phi?

 

[turin]

First try writing the Lagrangian in terms of x,y components, and you can use the freshman physics kinematic equations to include the contribution from the acceleration. However, I'm not suggesting that this should be the end result, but it may be the easiest way to see how the elevator motion and position enters into the Lagrangian and the constraint from the rod to get you started.

 

[flatmaster]

You shouldn't need to go to x and y. I assume you're allowed to have a non-inertial reference frame, so SImply add terms to U and T to account for the elevator motion

 

[flatmaster]

The height of the elevator assuming initial height and velocity of zero is

h=(a(t^2))/2
Where "a" is the elevator's acceleration

Thus
U=mgl(1-cos) + (mga(t^2))/2

u = mg{(1-cos) + at^2/2}

As you can see, the component in brackets is the actual height accounting for both motions.


Now you also have an additional velocity component ...

 

[flatmaster]

Velocity as a function of time for an acceleration a is...

v = a t

You need to be carefull when doing kenitic energy this way. Technically, you would need to find a final velocity vector combining both elevator and pendulum motion, then square that velocity. However, we assume SMALL ANGLE, thus, these two velocities are orthogonal and it works out.


For the kenitic energy because of the elevator's motion...
Ke = (1/2) m (at)^2

 

[flatmaster]

Combining to get total kinetic energy, you can see...
T = (1/2) m {(l phidot)^2 + (at)^2}

As you can see, the quantity in curly brackets is simply the square of the TOTAL velocity vector

 

[turin]

u = mg{(1-cos) + at^2/2}

As you can see, the component in brackets is the actual height accounting for both motions.

I assume that the missing "l" is just a typo. It's too bad we didn't get to see Jumblebee figure this one out.

 

[turin]

... we assume SMALL ANGLE, ...

Who's "we"? I don't. I didn't see that in the problem statement.

 

[leonne]

so the Lagrangian equation of motion would be -mgl(1+sin phi)=ml^2phi 2dots right?