In the absence of torque,the rotational frequ | vedclass

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Question

(D) In the absence of external torque,the angular momentum $L$ of the body remains conserved.$L = I \omega = 	ext{constant}$,where $I$ is the moment

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) In the absence of external torque,the angular momentum $L$ of the body remains conserved.$L = I \omega = 	ext{constant}$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.Since $\omega = 2\pi f$ (where $f$ is the frequency),$I_1 f_1 = I_2 f_2$.The moment of inertia $I$ is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.Substituting this into the conservation equation: $M k_1^2 f_1 = M k_2^2 f_2$.Since the mass $M$ is constant,$k_1^2 f_1 = k_2^2 f_2$.Therefore,$\frac{k_1^2}{k_2^2} = \frac{f_2}{f_1}$.Given $f_1 = 1 	ext{ cy/sec}$ and $f_2 = 16 	ext{ cy/sec}$,we have $\frac{k_1^2}{k_2^2} = \frac{16}{1}$.Taking the square root on both sides,$\frac{k_1}{k_2} = \sqrt{\frac{16}{1}} = \frac{4}{1}$.Thus,the ratio of the radius of gyration is $4 : 1$.

In the absence of torque,the rotational frequency of a body changes from $1 \text{ cy/sec}$ to $16 \text{ cy/sec}$. The ratio of the radius of gyration in the two cases will be: (in $: 1$)

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