Question:
Because of the friction between the water in oceans with the earth's surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by $0.0016$ rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 $\mathrm{km}$ and its mass is $6.0 \times 10^{24} \mathrm{~kg}$.
Solution:
$\alpha=\frac{\Delta \omega}{t}$
$=\frac{0.0016}{(86400 \times 100 \times 365 \times 86400)}$
$\tau=I \alpha$
$=\left(\frac{2}{5} M R^{2}\right) \alpha$
$=\frac{2}{5}\left(6 \times 10^{24}\right)(6400 \times 1000)^{2}\left(\frac{0.016}{(86400)^{2} \times 100 \times 365}\right)$
$\tau=5.8 \times 10^{20} \mathrm{~N}-\mathrm{m}$