Angular Momentum of the earth at the equator

1. Oct 20, 2008

vesperaka

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

For a 98 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation? Assume the Earth has a radius of 6.4 x 106 m.

2. Relevant equations

Angular Momentum = rmv * sin(theta)

3. The attempt at a solution

r= 6.4*10^6 m
m = 98kg
Vangular (of earth) = 7.27*10^5 rad/s

I multiplied these 3 values together and got 45597.44 kgm^2/s as my answer but it's wrong. Do I have to incorporate anything about the sin of the angle into my equation (I figured it would just be sin(90) = 1 though...)? I don't know what else to do :\

2. Oct 21, 2008

alphysicist

Hi vesperaka,

This number does not look right to me. You have units of radians per second, and so the angular velocity of the earth is 7.27*10^(-5) rad/s.

But this is not the v that is in your equation (it does not have the same units, for example). So you need to either find the speed v from this angular velocity, or use an equation for the angular momentum that has the angular velocity in it.