Question

A mass $m$ on a frictionless table is attached to a hanging mass by a spring (of spring constant $K$ and natural length $l_0$) and a cord through a hole in the table. If the hanging mass remains stationary and the mass $m$ moves on a circular path with angular velocity $\omega$ the radius of its circular path is

Answer 1

$\frac{Kl_0 - Mg}{K - m \omega^2}$

Answer 2

The mass $m$ is undergoing circular motion with angular velocity $\omega$ at a radius $r$. The forces acting on mass $m$ in the radial direction are the tension in the spring and the tension in the cord. Both forces are directed towards the center of the circular path.

The tension in the cord is equal to the weight of the hanging mass, which is $T_c = Mg$. This force acts radially inwards.

The spring has a natural length $l_0$ and spring constant $K$. When the mass $m$ is at a distance $r$ from the center, the extension or compression of the spring is $(r - l_0)$. If $r > l_0$, the spring is stretched, and it exerts an inward force $T_s = K(r - l_0)$. If $r < l_0$, the spring is compressed, and it exerts an outward force $T_s = K(l_0 - r)$.

For circular motion, the net radial force must provide the centripetal force, which is $F_c = m \omega^2 r$.

Case 1: Spring is stretched ($r > l_0$). The net inward force is $T_s + T_c = K(r - l_0) + Mg$. Equating this to the centripetal force: $K(r - l_0) + Mg = m \omega^2 r$ $Kr - Kl_0 + Mg = m \omega^2 r$ Rearranging the terms to solve for $r$: $Kr - m \omega^2 r = Kl_0 - Mg$ $r(K - m \omega^2) = Kl_0 - Mg$

Case 2: Spring is compressed ($r < l_0$). The net inward force is $T_c - T_s = Mg - K(l_0 - r)$. Equating this to the centripetal force: $Mg - K(l_0 - r) = m \omega^2 r$ $Mg - Kl_0 + Kr = m \omega^2 r$ Rearranging the terms to solve for $r$: $Kr - m \omega^2 r = Kl_0 - Mg$ $r(K - m \omega^2) = Kl_0 - Mg$

In both cases, the equation for $r$ is the same: $r(K - m \omega^2) = Kl_0 - Mg$

If $K - m \omega^2 \neq 0$, we can solve for $r$: $r = \frac{Kl_0 - Mg}{K - m \omega^2}$

This expression can also be written as: $r = \frac{-(Mg - Kl_0)}{-(m \omega^2 - K)} = \frac{Mg - Kl_0}{m \omega^2 - K}$

The final answer is $\frac{Kl_0 - Mg}{K - m \omega^2}$.