Question

If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ______ hours 30 minutes.

Answer 1

By conservation of angular momentum:

$I_1 \omega_1 = I_2 \omega_2$

Moment of inertia of a sphere:

$I = \frac{2}{5} M R^2$

Angular velocity relation:

$\omega = \frac{2 \pi}{T}$

$\frac{2}{5} M R^2 \cdot \frac{2 \pi}{T_1} = \frac{2}{5} M \left( \frac{3}{4} R \right)^2 \cdot \frac{2 \pi}{T_2}$

Simplifying:

$\frac{1}{T_1} = \frac{9}{16} \cdot \frac{1}{T_2}$

$T_2 = \frac{16}{9} \cdot T_1$

Substituting $T_1 = 24$ hours:

$T_2 = \frac{9}{16} \cdot 24 = \frac{27}{2} = 13 \, \text{hours} \, 30 \, \text{minutes}.$

Answer 2

By conservation of angular momentum:

$I_1 \omega_1 = I_2 \omega_2$

Moment of inertia of a sphere:

$I = \frac{2}{5} M R^2$

Angular velocity relation:

$\omega = \frac{2 \pi}{T}$

$\frac{2}{5} M R^2 \cdot \frac{2 \pi}{T_1} = \frac{2}{5} M \left( \frac{3}{4} R \right)^2 \cdot \frac{2 \pi}{T_2}$

Simplifying:

$\frac{1}{T_1} = \frac{9}{16} \cdot \frac{1}{T_2}$

$T_2 = \frac{16}{9} \cdot T_1$

Substituting $T_1 = 24$ hours:

$T_2 = \frac{9}{16} \cdot 24 = \frac{27}{2} = 13 \, \text{hours} \, 30 \, \text{minutes}.$