Conservation of angular momentum of a falling particle

1. Jan 7, 2010

Rudipoo

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

A stone is dropped from a stationary helicopter 500m above the ground, at the equator. How far from the point vertically below the helicopter does it land?

2. Relevant equations

Conversation of AM

3. The attempt at a solution

Let the height above the ground it is dropped be h, the radius of the Earth R, the mass of the stone m (which will cancel) and the angular velocity of the earth w. Then the angular momentum as it's dropped is mw(R+h)^2.

When the particle is at a height y above the Earth's surface, the stone has angular momentum m(R+y)(v_x+v_0) where v_0=(R+h)w is the velocity (in x direction) when it dropped, due to the helicopter being stationary, w.r.t. the Earth.

Now, y=h-0.5g*t^2, and Conservation of AM implies

mw(R+h)^2=m(R+h-0.5g*t^2)(v_x+(R+h)w).

I rearranged for v_x and integrated between t=0 and t'=Sqrt(2h/g), the time for the stone to hit the ground.

I go the answer x=12cm, but it should be x=24cm. Am i performing the integration incorrectly or have I set up the equations wrong?

Thanks, Rupe

2. Jan 7, 2010

diazona

I get the same answer of 12cm by doing the integral, so it seems like the problem is in your setup. Though I don't immediately see what you might have done incorrectly.

3. Jan 11, 2010

Rudipoo

Thanks for your help diazona. That improves my confidence in the integration. I wonder if anyone can see my mistake in setting up the equation?

Thanks.