As shown in the figure, a ring attached with a spring is fitted in a smooth rod. The spring is fixed

Question

As shown in the figure, a ring attached with a spring is fitted in a smooth rod. The spring is fixed at the outer end of the rod. The mass of the ring is $3 	ext{ kg}$ & spring constant of spring is $300 	ext{ N/m}$. The ring is given a velocity $V$ towards the outer end of the rod. And the rod is set to be rotating with an angular velocity $\omega$. The natural length of the spring is same as that of length of rod. Then ring will move with constant speed with respect to the rod if:

Professor PC Thomas Classes · Answer

Let rod length be $L$. If the ring is at distance $x$ from the axis, the spring (fixed at the outer end) has length $L-x$. Its natural length is $L$, so it is compressed by $x$. The spring force is $kx$ directed inward. The centrifugal force is $m\omega^2x$ directed outward. For constant speed in the rod's frame, net force must be zero everywhere: $m\omega^2x - kx = 0 \implies \omega = \sqrt{k/m}$. With $k = 300 	ext{ N/m}$ and $m = 3 	ext{ kg}$, $\omega = \sqrt{300/3} = 10 	ext{ rad/s}$.