Question
Without any torque, the angular velocity of an object change from 1 revolution per 25 second to 1 revolution per second. Find the ratio of the radii of rotation of the object in these two situations
✓
Answer
We know
From the law of conservation of angular momentum
$I_1 \omega_1=I_2 \omega_2$
or, $\quad I _1 \times 2 \pi n_1= I _2 \times 2 \pi n_2$
when $n_1$ and $n_2$ are frequencies,$
I_1 n_1=I_2 n_2$
If the radii of rotation in both the positions of the body are $K _1$ and $K _2$ in that situation
$MK _1^2 n_1= MK _2^2 n_2$
$\therefore \quad k_1^2 n _1=k_2^2 n_2$
$\frac{ K _1^2}{K_2^2}=\frac{n_2}{n_1}$
Given $n_1=\frac{1}{25}, n_2=1$
$\frac{ K _1^2}{K_2^2}=\frac{1}{\frac{1}{25}}=\frac{25}{1}$
$\frac{ K _1}{K_2}=\sqrt{\frac{25}{1}}=\frac{5}{1}$
$\frac{ K _1}{K_2}=\frac{5}{1}$ Ans.
From the law of conservation of angular momentum
$I_1 \omega_1=I_2 \omega_2$
or, $\quad I _1 \times 2 \pi n_1= I _2 \times 2 \pi n_2$
when $n_1$ and $n_2$ are frequencies,$
I_1 n_1=I_2 n_2$
If the radii of rotation in both the positions of the body are $K _1$ and $K _2$ in that situation
$MK _1^2 n_1= MK _2^2 n_2$
$\therefore \quad k_1^2 n _1=k_2^2 n_2$
$\frac{ K _1^2}{K_2^2}=\frac{n_2}{n_1}$
Given $n_1=\frac{1}{25}, n_2=1$
$\frac{ K _1^2}{K_2^2}=\frac{1}{\frac{1}{25}}=\frac{25}{1}$
$\frac{ K _1}{K_2}=\sqrt{\frac{25}{1}}=\frac{5}{1}$
$\frac{ K _1}{K_2}=\frac{5}{1}$ Ans.
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