Question

Determine the speed with the earth would have to rotate on its axis so that a person on the equator would weight 3/5 as much as present

Answer 1

True weight at equator is w=mg and observed weight is W' = mg' = 3/5 mg
taking λ= 0.
    mg'      = mg - mRw^2 cosλ
    3/5mg = mg - mRw^2 cos0
                = mg - mRw^2
or
  mRw^2 = 2/5 mg
  
w = (2g/5R)^1/2
    = 7.8 * 10^-4 rad/sec 

Answer 2

Here we have to determine the speed with the earth would have to rotate on its axis so that a person on the equator would weight 3/5 as much as present.
Solution:
Actual weight on the equator = W = mg
where 'm' is the mass and 'g' is the gravity.
According to the given condition,
Weight on the equator = W' = mg'

Weight on the equator = W' = 3/5 mg   ........... (1)
We know that, λ = 0 at the equator.
Now,
mg' = mg - m R ω² cos  λ
3/5 mg = mg - m R ω² cos 0   (∵ λ = 0 )
3/5 mg = mg - m R ω² (1)      (∵ cos 0 =1)
3/5 mg = mg - m R ω²
m R ω² = mg - 3/5 mg
m R ω² = (1 - 3/5) mg
m R ω² = 2/5 mg
R ω² = 2/5 g
ω² = 2 g / 5 R
ω² = (2 x 10) / (5 x 6.4x10⁶)   (∵ R = Radius of Earth = 6400 km = 6.4x10⁶ m)
ω = √ (6.25x10⁻⁷)
ω = 7.8 x 10⁻⁴ rad/sec
This is the required answer.
Thanks.