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 earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case ?

Answer 1

(a) Net force on the spring balance,

$R=mg=m{\omega }^{2}r$

So, fraction less than the true weight (3mg) is

$=\frac{mg-(mg-m{\omega }^{2}r)}{mg}$

$=\frac{{\omega }^{2}r}{g}{\left(\frac{2\pi }{24\times 3600}\right)}^2$

$=\frac{6500\times {10}^3}{10}$

(b) When the balance reading is half the true weight,

$\frac{mg-m{\omega }^{2}r}{mg}=\frac{1}{2}$

$\Rightarrow {\omega }^{2}r=\frac{g}{2}$

$\Rightarrow \omega =\sqrt{\frac{g}{2\pi }}$

$=\sqrt{\frac{10}{2\times 6400\times {10}^3}}rad/sec$

$\therefore$ Duration of the day is

$T=\frac{{}^{\prime }2\pi }{\omega }$

$=2\pi \times \sqrt{\frac{2\times 6400\times {10}^3}{9.8}}sec$

$=2\pi \sqrt{\frac{64\times {10}^5}{49}}sec$

$=\frac{2\pi \times 8000}{7\times 3600}hr=2hr$