For a 99 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation?

L = 4.58 x 10⁴ kg.m²/s

Explanation:

The angular momentum is given by the formula:

L = mvr

but, v = rω

Therefore,

L = mr²ω

where,

L = Angular Momentum of the person = ?

m = mass of person = 99 kg

r = radius of earth = 6.37 x 10⁶ m

ω = Angular Speed of Earth's Rotation = θ/t

Since, earth completes 1 rotation in 1 day. Hence,

ω = (2π rad/1 day)(1 day/24 h)(1 h/3600 s)

ω =  7.27 x 10⁻⁵ rad/s

Therefore,

L = (99 kg)(6.37 x 10⁶ m)²(7.27 x 10⁻⁵ rad/s)

L = 4.58 x 10⁴ kg.m²/s

6 cos 34, 6 sin 34 trig