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 km and its mass is 6.0 x 10²⁴ kg. ?

Answer 1

Given in the question :-

Angular speed is decrease by 0.0016 rad / day in 100 years.

Time t = 100 yrs.

or time in terms of day t, = 100 × 365  = 36500 day.

ω(Initial) = 2 π rad /day

ω'(final) = 2 π - 0.0016 rad /day

Now, we have to find $\alpha$ = ?

Now we know the formula ,

$\omega' = \omega - \alpha t$

Now put the given value in this formula

2 π- 0.0016 = 2 π - α × 36500.

$\alpha = \frac{0.0016}{36500}$

$\alpha = 4.4 * 10^-^8 rad/day^2$

α = 4.4 x 10⁻⁸ × (24 × 3600²) [on changing into sec]

$\alpha = 5.87 * 10^-^1^8 rad/s^2$

Here given that the Moment of inertia of the earth

$= \frac{2}{5} MR^2$

= 0.40 × 6.0 × 10²⁴ × 6400000²

= 9.8 x 10³⁷

Average torque produce by the ocean of the friction force = Iα

= 9.83 x 10³⁷ × 5.87 x 10⁻¹⁸

=$\boxed{5.77 * 10^2^0 N.m}$

Hope it Helps :-)