Question

A person stands on a spring balance at the equator.
(a) By what friction is the dinbalance reag less than hits 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) Balance reading = Normal force on the balance by the Earth.
At equator, the normal force (N) on the spring balance:
N = mg − mω²r

True weight = mg
Therefore, we have:
$$\begin{gathered} \mathrm{Fraction}\mathrm{less}\mathrm{than}\mathrm{the}\mathrm{true}\mathrm{weight}=\frac{mg-(mg-m{\mathrm{\omega }}^{2}r)}{mg} \\ =\frac{{\mathrm{\omega }}^{2}r}{g}={\left(\frac{2\mathrm{\pi }}{24\times 3600}\right)}^2\left(\frac{6.4\times {10}^6}{10}\right) \end{gathered}$$
$=3.5\times {10}^{-3}$

(b) When the balance reading is half, we have:
$\mathrm{True}weight=\frac{mg-m{\mathrm{\omega }}^{2}r}{mg}=\frac{1}{2}$



$$\begin{gathered} \Rightarrow {\mathrm{\omega }}^{2}r=\frac{g}{2} \\ \Rightarrow \mathrm{\omega }=\sqrt{\frac{g}{2r}} \\ =\sqrt{\frac{10}{2\times 6400\times {10}^3}}\mathrm{rad}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Duration}\mathrm{of}\mathrm{the}\mathrm{day}=2\pi \times \sqrt{\frac{2\times 6400\times {10}^3}{9.8}}\mathrm{s} \\ =2\mathrm{\pi }\sqrt{\frac{6.4\times {10}^7}{49}}\mathrm{s} \\ =\frac{2\mathrm{\pi }\times 8000}{7\times 3600}\mathrm{h}=2\mathrm{h} \end{gathered}$$