# Mechanics Problem (falling object)

1. Aug 15, 2012

### infke

Hello, let me start with saying this is not homework but rather a question I have asked myself and cannot seem to solve.

1. The problem statement, all variables and given/known data
Basically this:

If a ladder is standing against a wall (both in a 90° corner with the ground) and falls over, what is the approximate speed of the highest point of the ladder as it hits the ground? The only force you should take note of is the gravitational acceleration. Because of the state of the ladder it shouldn't move according to the laws of physics but let's assume you give it a neglectable push to start the movement. You can also assume that the part touching the ground does not move.

The problem with this is that you don't have a linear acceleration. It varies anywhere from 0 to g (9.81m/s²) and I have no idea how to solve this correctly.

2. Relevant equations
The distance (x) the highest point has traveled over is
x = r * ∏/2
with r being the length of the ladder

Also the acceleration along the y curve on any given point A on the x curve smaller than r
a = g * cos ( A / r )

3. The attempt at a solution
Because the acceleration is not linear I did not find a correct way to calculate my question.

Last edited: Aug 15, 2012
2. Aug 15, 2012

### Nessdude14

This problem requires some calculus (integration) if you want to find the final velocity of the ladder. I'll tell you how to find how fast the ladder is accelerating for a given angle, and if you want, we can go on to the integration from there.

You can find the torque at the ladder's pivot point by using the equation:
$τ=r_{\perp}F$ or $τ=rFsinϕ$
Where ϕ is the angle between the gravity vector and the axis along the ladder's length.
You can assume that the force is being applied at the ladder's center of mass (so the length of r is equal to half the ladder's height).

Using Newton's second law for rotation, we can use this torque to determine the angular acceleration:
$\large α=\frac{τ}{I}$
I is the ladder's moment of inertia.
If we assume the ladder to be a thin rod, the moment of inertia for rotation about the end is:
$I=\frac{1}{3}ML^{2}$

If we substitute τ and I into the rotation equation, we get:
$\large α=\frac{3rFsinϕ}{ML^{2}}$
We can make two more substitutions: F=Mg and r=L/2
$\large α=\frac{3}{2}\frac{LMgsinϕ}{ML^{2}}$
Now we can cancel out an L and M from both sides of the fraction.
$\large α=\frac{3}{2}\frac{gsinϕ}{L}$
Since ϕ is a bit of an odd angle, let's substitute ϕ=θ+90°, making θ the angle between the ground and the ladder.
$\large α=\frac{3}{2}\frac{gsin(θ+90)}{L}$
and we can substitute sin(θ+90)=cosθ to simplify it a bit more:
$\large α=\frac{3}{2}\frac{gcos(θ)}{L}$
This equation gives the angular acceleration for a ladder of length L at an angle θ from the ground. Since the angle in this equation is a function of time, this is where you'd need calculus to move on with the problem.

Last edited: Aug 15, 2012
3. Aug 15, 2012

### azizlwl

Using conservation of energy, PE=KE
PE=mgh
$PE=g\int_0^d \! x \, \mathrm{d} m$
KE=0.5Iω2