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(A) Since no external torque is mentioned to be acting on the system in a way that changes the angular momentum,we assume the angular momentum $L$ rem

Vedclass · Accepted Answer

$\sqrt{\omega_2} : \sqrt{\omega_1}$

Vedclass · Answer

(A) Since no external torque is mentioned to be acting on the system in a way that changes the angular momentum,we assume the angular momentum $L$ remains constant $(L_1 = L_2)$.Angular momentum is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.Thus,$I_1 \omega_1 = I_2 \omega_2$.The moment of inertia is $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration.Substituting this,we get $M K_1^2 \omega_1 = M K_2^2 \omega_2$.Canceling $M$ from both sides: $K_1^2 \omega_1 = K_2^2 \omega_2$.Rearranging for the ratio of the radii of gyration: $\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$.Taking the square root of both sides: $\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}} = \frac{\sqrt{\omega_2}}{\sqrt{\omega_1}}$.Therefore,the ratio is $\sqrt{\omega_2} : \sqrt{\omega_1}$.

$A$ torque is applied to a body,changing its angular velocity from $\omega_1$ to $\omega_2$. What is the ratio of the initial radius of gyration to the final radius of gyration?

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