Ladder leaning on frictionless wall and frictionless floor

[WackStr]

Homework Statement

A ladder of length L is leaning against a frictionless wall. The floor is frictionless too. It starts to fall. At what angle will it leave the wall.

Homework Equations

The Attempt at a Solution

I can derive the equation of motion, \ddot{\theta}=\frac{3}{2}\frac{g}{L}\cos\theta . I just can't find the angular acceleration at which the ladder should leave the wall. I think there is some simple reasoning involved. Once I find it the problem is solved. Need help with finding that value of \ddot{\theta}

 

[Quinzio]

There is not a fixed angle. It depends on the initial position of the ladder, where you leave it free to fall.

Image you find a fixed angle. Let's say 40°. If I position the ladder at 35°, and leave it free, in that moment it touches the wall, and I bet even shortly after.

You end up with the angle being a function of the initial position and it's a solution of a differential equation. There' s not a simple shortcut.

 

[WackStr]

There is not a fixed angle. It depends on the initial position of the ladder, where you leave it free to fall.

Image you find a fixed angle. Let's say 40°. If I position the ladder at 35°, and leave it free, in that moment it touches the wall, and I bet even shortly after.

You end up with the angle being a function of the initial position and it's a solution of a differential equation. There' s not a simple shortcut.

Oh yea sorry ... I missed that part. We do know the initial angle. My question was what is the physical condition that determines the angle that the ladder flys off given the initial angle. (that is, what mathematical constraint do we need in conjunction with the equation of motion to determine this angle?

 

[Quinzio]

Is there any advance for this problem ? It would be interesting to share the solutions.
I've come to a method, but I'm not sure about it.

 

[WackStr]

I think that I had a problem a year or two back where I think we solved this problem. However, I think I must be mistaken because the equation of motion cannot be analytically solved.

I don't have the solution.

I still am thinking about what would happen when the ladder leaves the wall (I know the normal force goes to zero, but how do we translate it into a constraint on \theta and thus solve the differential equation and get the answer.)

 

[Dmytry]

Simple reasoning involved all right. Consider horizontal momentum. That is, let x be horizontal position of centre of mass, when \ddot{x}<0 , ladder leaves the wall.

 

[Dmytry]

Interesting idea.
But what about the rotation of the ladder ?
It could keep the ladder in contact to the wall even if the CoM is not accelerating anymore.

Seems harder than that. Maybe I'm wrong.

Consider conservation of linear momentum. The horizontal momentum of the ladder equals V_cm*M , regardless of the angular momentum. Linear momentum conserves independently from angular momentum, and the
\sum \vec{F}=\ddot{\vec{X}} _{cm}*m
equation holds regardless of whenever the forces also make the object rotate.
The only horizontal force is by ladder touching the wall.

 

[Quinzio]

Thanks for the answers. As this problem is intriguing, I will try to reach the end, although I still think it's not that easy. For reasons I explained above, the "leaving" angle if a function of the initial angle, where we suppose the ladder is still and left to fall.

We have a ladder:

L is the length

Y the contact point on the wall

X the contact point on the floor

V_X the speed of X

V_Y the speed of Y

V_{CM} the speed of the CoM

\theta the angle over the horizontal

\omega the angular speed

As long as the ladder touches the wall, the CoM moves along a quarter of circle, and simplifies things a great deal.

We notice soon that:
X = \sqrt{L^2-Y^2}

and
\theta = - arctg { \left ( Y \over X \right )}

I'll give \theta as a function of Y

\theta = - arctg \ {Y \over X} = - arctg \left({Y \over {\sqrt{L^2-Y^2}}}\right)

derivating {d \theta \over dY} = - {1 \over \sqrt{L^2-Y^2}}

by which we get

\omega = - {V_Y \over \sqrt{L^2-Y^2}}

The total energy of the body is:

{1 \over 2} m {V_Y}^2 + { 1\over 6 } m L^2 \omega ^2

I'll simplify the mass to 1

E ={1 \over 2} {V_Y}^2 + { 1\over 6 } L^2 \omega ^2

E ={1 \over 2} {V_Y}^2 + { 1\over 6 } L^2 {V^2_Y \over {L^2-Y^2}} = <br /> V^2_Y \left( {4L^2-3Y^2 \over 6L^2-6Y^2 } \right)

Since the contact with the wall and floor is frictionless, no energy is lost, and so all te potential energy lost by the CoM goes into kinetic energy.

If Y_0 is the initial wall contact point, and m=1 .
E = mgh = g \left({Y_0-Y \over 2} \right)
g \left({Y_0-Y \over 2} \right) = V^2_Y \left( {4L^2-3Y^2 \over 6L^2-6Y^2 } \right)

V_Y = \sqrt {g (Y_0-Y) (6L^2-6Y^2) \over 2( 4L^2-3Y^2) }