Question

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

• A. $1.3\times 10^{22}N$
• B. $8.26\times 10^{28}N$
• C. $1.3\times 10^{23}N$
• D. None of these

Answer 1

Answer: (A)

${\omega }_1=2\pi$ rad / day, ${\omega }_2=0$ and $t=1$ day
$\therefore \alpha =\frac{{\omega }_2-{\omega }_1}{t}=\frac{0-2\pi }{1}$
$\frac{rad}{da{y}^2}\frac{2\pi }{(86400{)}^2}\frac{rad}{s^2}$
Torque required to stop the earth, $\tau =l\alpha =FR$
$F=\frac{l\alpha }{R}=\frac{\frac{2}{5}M{R}^2\times \alpha }{R}=\frac{2}{5}MR\times \alpha$
$=\frac{2}{5}\times 6\times 10^{24}\times 6400\times 10^3\times \frac{2\pi }{(86400{)}^2}$
$=1.3\times 10^{22}N$