Question

please solve this question
How this relation came?​

Answer 1

Answer:

All you need to do is apply rotational energy conservation . I.e.

$\frac{1}{2} I { \omega}^{2} = const.$

Now we need ratio of mass and radius for momentum of inertia .

So Given that

48.8 water leaks or 51.2 % remains , so

$\frac{m_2}{m_1} = \frac{51.2}{100} = \frac{512}{1000}$

Now note that mass = density × volume , and volume is proportional to cube of radius so

$\frac{m_2}{m_1} = (\frac{r_2}{r_1} ) {}^{3} = \frac{512}{1000} \\ \\ \implies\frac{r_2}{r_1} = \sqrt[3]{\frac{512}{1000}} = \frac{8}{10} = \frac{4}{5}$

So now

$\frac{1}{2} I_1 { \omega_1}^{2} = \frac{1}{2} I _2{ \omega_2}^{2} \\ \\ \frac{2}{5} m _1{r_1}^{2} \times { \omega_1}^{2} = \frac{2}{5} m _2{r_2}^{2}{ \omega_2}^{2} \\ \\ { \omega_2}^{2} = \frac{m_1}{m_2} \times (\frac{r_1}{r_2}) {}^{2} \times { \omega_2}^{2} \\ \\ { \omega_2}^{2} = \frac{1000}{512} \times ( \frac{4}{5} ) {}^{2} \times { 60}^{2} \\ \\ { \omega_2}^{2} = \frac{5}{4} \times {60}^{2} \\ \\ { \omega_2} = 30 \sqrt{5} \: rpm$