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Circular Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsCircular Motion5 Marks
Question
A person stands on a spring balance at the equator.
  1. By what fraction is the balance reading less than his true weight?
  2. 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

When the balance reading is half, we have:$=\frac{\text{mg}-\text{m}\omega^3\text{r}}{\text{mg}}-=\frac{1}{2}$
True weight =$\Rightarrow\omega^2\text{r}=\frac{\text{g}}{2}$
$\Rightarrow\omega=\sqrt{\frac{\text{g}}{2\text{r}}}$
$=\sqrt{\frac{12}{2\times6400\times10}}\text{rad}/\text{ s}$
$\therefore\text{Duration of the day}=2\pi\times\sqrt{\frac{2\times6400\times10^2}{9.8}}\text{s}$
$=2\pi\sqrt\frac{{6.4\times10^7}}{49}\text{s}$
$=\frac{2\pi\times8000}{7\times3600}\text{h}=2\text{h}$

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