Angular Momentum of the earth at the equator

[vesperaka]

Homework Statement

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.

Homework Equations

Angular Momentum = rmv * sin(theta)

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 :\

 

[alphysicist]

Hi vesperaka,

Homework Statement

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.

Homework Equations

Angular Momentum = rmv * sin(theta)

The Attempt at a Solution

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

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.