Question

A spherical water balloon is revolving at 60rpm . In the course of time . 48.8% of it's water leaks out. With what frequency will the remaining ballon revolve now ? Neglect all non conservative force?​

Answer 1

We know that,

$\frac{m_1}{m_2} = \frac{v_1}{v_2} = \frac{r_1}{r_2}$ [∵ all non-conservative forces are neglected]

∴ Linear momentum will remain constant

⇒ mvr will remain constant

But, according to the question,

the object is a spherical water balloon; then $r_{3}$ is used

So,

$\frac{r_1}{r_2} = (\frac{m_1}{m_2})$ × $\frac{1}{3}$

Also, we have $\frac{m_1}{m_2}$ = $\frac{100}{100-48.4}$ = $\frac{100}{51.2}$ = $\frac{1}{0.512}$

Then,  $(\frac{m_1}{m_2})^{\frac{1}{3} } = 1.25$

So, the moment of inertia of the spherical water balloon = $\frac{2}{5} mr_{2}$

Then,

$\frac{m_1}{m_2}$ × $\frac{r_1}{r_2}^{2} = (\frac{m_1}{m_2} )^{\frac{5}{3} }$

∴ With the help of conservation of linear momentum we obtain I1W1 =I2W2

Ans) 3.052 rps

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