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  4. What constant force tangential to the equator should be applied to the earth to stop its rotation is one day?

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Q. What constant force tangential to the equator should be applied to the earth to stop its rotation is one day?

System of Particles and Rotational Motion

Solution:

$\omega_{1}=2\, \pi$ rad / day, $\omega_{2}=0$ and $t=1$ day
$\therefore \alpha=\frac{\omega_{2}-\omega_{1}}{t}=\frac{0-2 \pi}{1}$
$\frac{rad}{day^{2}} \frac{2 \pi}{(86400)^{2}} \frac{r a d}{s^{2}}$
Torque required to stop the earth, $\tau=l \alpha=F R$
$F=\frac{l \alpha}{R} =\frac{\frac{2}{5} M R^{2} \times \alpha}{R}=\frac{2}{5} M R \times \alpha$
$=\frac{2}{5} \times 6 \times 10^{24} \times 6400 \times 10^{3} \times \frac{2 \pi}{(86400)^{2}}$
$=1.3 \times 10^{22} N$