Moment of intertia, lagrangian, etc

[Shafikae]

A uniform ladder of mass M and length 2L is leaning aainst a frictionless vertical wall with its feet on a frictionless horizontal floor. Initially the stationary ladder is released at an angle $$\theta_{0}$$ = 60^o to the floor. Assume that the gravitation field g acts vertically downward.

1) Show that the moment of intertia of the ladder is about its midpoint is I = (1/3)ML²

2) Derive the Lagrangian of the system.

3) Derive the equations of motions using the Lagrangian.

4) Find the angle $$\theta$$ at which the ladder loses contact with the vertical wall.


Ive been reading my textbooks but there arent any similar examples for me to relate to.

 

[jdwood983]

A uniform ladder of mass M and length 2L is leaning aainst a frictionless vertical wall with its feet on a frictionless horizontal floor. Initially the stationary ladder is released at an angle $$\theta_{0}$$ = 60^o to the floor. Assume that the gravitation field g acts vertically downward.

1) Show that the moment of intertia of the ladder is about its midpoint is I = (1/3)ML²

2) Derive the Lagrangian of the system.

3) Derive the equations of motions using the Lagrangian.

4) Find the angle $$\theta$$ at which the ladder loses contact with the vertical wall.


Ive been reading my textbooks but there arent any similar examples for me to relate to.

For (1), you need to calculate the moment of inertia or a rod: $I=cmr^2$ where c is some number. Methods of calculating the moment of inertia of any object should be found in your 1st year physics textbooks.
For (2), you have a constant-length rod that is falling from some initial angle. Sounds like it would require rotational kinetic energy, $T=\frac{1}{2}I\omega^2$, added to the potential energy for your Lagrangian: $L=\frac{1}{2}I\omega^2+V$.
For (3) and (4), once you have (2), you should be able to do these two.

 

[Shafikae]

I was able to obtain part 1.
and concerning part 2, the potential would just be MgL or in our case 2MgL?

 

[jdwood983]

I was able to obtain part 1.
and concerning part 2, the potential would just be MgL or in our case 2MgL?

While the latter has the appropriate form, you are not taking into account the angle for which this ladder is leaning. Multiply the above by the appropriate trigonometric function, and there's your potential.

 

[Shafikae]

I don't understand. Do u mean that the potential V = 2MgLSin60 ?

 

[jdwood983]

I don't understand. Do u mean that the potential V = 2MgLSin60 ?

I wouldn't necessarily put $\theta_0$ in there, I would use just plain $\theta$. As for the actual trigonometric function, it completely depends on where the angle is being measured, but it would be either cosine or sine.

 

[Shafikae]

and would it be negative potential since the gravity is acting vertically downward?

 

[Shafikae]

and what is meant by the equations of motion?

 

[jdwood983]

and what is meant by the equations of motion?

How far into Lagrangian mechanics have you studied in class? The equations of motion are the reason for utilizing Lagrangian mechanics. Have you seen this equation:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)-\frac{\partial L}{\partial q}=0$$

 

[Shafikae]

yes I've seen that before.

 

[jdwood983]

Well if you let $\dot{q}=\dot[\theta]$ and $q=\theta$, you take the respective derivatives of your Lagrangian and find that you will have something of the order

$$m\ddot{q}=f(q,\dot{q},t)$$

where $f(q,\dot{q},t)$ is something that will depend on coordinates, velocities, and possibly time. This is your equation of motion.

 

[Shafikae]

so its only in terms of $$\dot{\theta}$$ ? we don't do it also in terms of the length of the rod/ladder?

 

[jdwood983]

so its only in terms of $$\dot{\theta}$$ ? we don't do it also in terms of the length of the rod/ladder?

More than likely those constants will be factors in that right hand side, however the more important factor in determining the acceleration would be the position and velocity of the rod, that is $\theta$ and $$\dot{\theta}$$.

Also, I don't think I mentioned this part, but in the kinetic energy term, $$\omega=\dot{\theta}$$. If you don't set these equal to each other, you won't get the answer necessary.

 

[Shafikae]

ahhhh ok got it! How do I determine if we will be using Cos or Sin? Mechanics and Classical Physics arent my thing, I am more interested in atomic and electronics

 

[jdwood983]

ahhhh ok got it! How do I determine if we will be using Cos or Sin? Mechanics and Classical Physics arent my thing, I am more interested in atomic and electronics

You'll determine which trig function to use by using trigonometry; it entirely depends on where your angle is being measured from.