Question

A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight? (b) If the speed of the earth rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of day in this case?

Answer 1

(a). We know the Formula,

g' = g - ω²r

∴  g - g' = ω²r

⇒ (g - g')/g = ω²r/g

∵ (g - g')/g is the fraction of weight which is less than its true weight.

⇒ Fraction of weight = ω²r/g

= 3.45 × 10⁻³

(b). In this case the outward force due to rotation of earth= Half the true weight  

mω²R=½mg

→2(2π/T)²R=g    

→8π²R=gT²

→T²=8π²R/g =8 x 9.86 x 6400 x 1000/9.8 = 51513469.4  

→T = 7177 s = 7177/3600  hr = 1.994 hr ≈ 2 hrs

Hope it helps.