Ladder sliding down with no friction

[Dustgil]

Homework Statement

A uniform ladder leans against a smooth vertical wall. If the floor is also smooth, and the initial angle between the floor and the ladder is theta(0), show that the ladder, in sliding down, will lose contact with the wall when the angle between the floor and the ladder is

sin^{-1}(\frac {2} {3} sin \theta_{0})

Homework Equations

\frac {dL} {dt}=N

where L is the angular momentum and N is the net external torque

The Attempt at a Solution

I understand why the ladder loses contact with the wall at some point. There is a normal force on the ladder exerted by the wall that is not balanced by anything. So as the ladder falls, it sort of gains length in the x direction while at the same time being pushed away from the wall. Eventually there comes a point where the normal force exerted by the wall is zero and then negative, which will be at the angle I'm looking for.

Let's say that the angle that the ladder makes with the floor is theta. The translational motion equations are

m \ddot y=F_{n} - mg
m \ddot x=F_{w}I start to get a bit confused when trying to define the rotational motion. The first question is where's the best place to define the rotation at? For now I choose the point where the ladder makes contact with the ground. This is where I run into my second problem. For whatever reason I'm having trouble confidently expressing the torques exerted from both gravity and the normal force from the wall. I'm just kinda getting mixed up by it. I make a guess:

\frac {dL} {dt}= N
I \ddot {\theta} = mgcos(\theta)(\frac {l} {2}) - F_{w}sin(\theta)(\frac {l} {2})

Are these the correct equations? It took me a long while to get to this point so It'd be cool to be confident before I soldier on.

 

[kuruman]

The first question is where's the best place to define the rotation at?

Try the corner of the wall and look at the center of mass. First understand the path it follows if both ends of the ladder are constrained to be always in contact, then figure out what must be true for the CM to deviate from this path.

 

[Dustgil]

So the CM moves in a circle. Won't the CM stop accelerating in the x direction when it loses contact?

 

[Dustgil]

That makes the translation motion

m \ddot {r} = r \dot { \theta }

but I'm still unsure as to how to define the rotational motion. maybe I need to brush up on my trig..

 

[kuruman]

So the CM moves in a circle.

Correct. What force provides the necessary centripetal acceleration? When that force goes to zero, the CM leaves the circle. Hint: This is essentially the same problem as the small block that slides down on the outside of a hemispherical frictionless surface.

 

[Dustgil]

For the hint it would be the contact force the block makes with the surface, so the normal force on the wall, right?

 

[kuruman]

For the hint it would be the contact force the block makes with the surface, so the normal force on the wall, right?

And the floor.

 

[haruspex]

Correct. What force provides the necessary centripetal acceleration? When that force goes to zero, the CM leaves the circle. Hint: This is essentially the same problem as the small block that slides down on the outside of a hemispherical frictionless surface.

Not sure that is going to work... Not obvious to me, anyway. The particle on the sphere has no moment of inertia.
The force "providing centripetal acceleration" is not one applied force, it is the radial component of the net force. Even though the normal force from the wall becomes zero, there is still gravity and the normal force from the floor. And this net component does not go to zero, it just becomes insufficient to maintain circular motion.

 

[kuruman]

And this net component does not go to zero, it just becomes insufficient to maintain circular motion.

I am not sure what "insufficient" means. The condition for the CM to go around in a circle is $$F_r -mg\cos \theta = - \frac{mv^2}{r} $$ where $F_r$ is the radial component of the contact force resultant. From this $$F_r = m\left(g\cos \theta - \frac{v^2}{r} \right)$$As long as $F_r$ is positive, circular motion of the CM is maintained. What is the criterion for "insufficient" other than $F_r = 0$?

 

[haruspex]

I am not sure what "insufficient" means. The condition for the CM to go around in a circle is $$F_r -mg\cos \theta = - \frac{mv^2}{r} $$ where $F_r$ is the radial component of the contact force resultant. From this $$F_r = m\left(g\cos \theta - \frac{v^2}{r} \right)$$As long as $F_r$ is positive, circular motion of the CM is maintained. What is the criterion for "insufficient" other than $F_r = 0$?

You wrote that the force necessary for centripetal acceleration goes to zero. That is not F_r; it is mg cos(θ)-F_r.

 

[kuruman]

Yes, thank you for the correction. I made the mistake of trying to do it in my head without writing anything down - I should know better.

 

[ehild]

Let's say that the angle that the ladder makes with the floor is theta. The translational motion equations are

m \ddot y=F_{n} - mg
m \ddot x=F_{w}

What are x and y? Are they the coordinates of the CoM? how are the related to the angle theta?
Yes, you can solve the problem by writing the equation of motion of the center of mass, and also the equation of rotation about the CoM.

One other approach can be using conservation of energy. The kinetic energy is the sum of the KE of the CoM, and the rotational energy about the CoM. As you have found, the CoM moves around a circle. That gives you a relation between θ and $\dotθ$.
At the end, when the ladder loses contact with the wall, $\ddot x$=0. You can write it also in terms of theta and derivatives.