Question

A small mass m is attached to a rubber cord and revolves in a horizontal frictionless plane with a constant frequency f. The cord lies in the plane of the circle. The unstretched length of the cord is $\ell_0$. The tension in the cord increases in direct proportion to its elongation, the tension per unit elongation being k. Find: (a) the radius of the uniform circular motion, (b) the tension T in the cord.

Answer 1

(a) The radius of the uniform circular motion is: $r = \frac{k\ell_0}{k - 4\pi^2 f^2 m}$

(b) The tension T in the cord is: $T = \frac{4\pi^2 f^2 m k\ell_0}{k - 4\pi^2 f^2 m}$

Answer 2

Let $r$ be the radius of the circular motion and $T$ be the tension in the cord. The elongation of the cord is $\Delta\ell = r - \ell_0$. According to Hooke's law for the rubber cord, the tension is given by $T = k \Delta\ell = k(r - \ell_0)$.

For uniform circular motion, the tension $T$ provides the necessary centripetal force $F_c$. The angular velocity is $\omega = 2\pi f$, where $f$ is the frequency. The centripetal acceleration is $a_c = \omega^2 r = (2\pi f)^2 r = 4\pi^2 f^2 r$. The centripetal force is $F_c = m a_c = m (4\pi^2 f^2 r)$.

Equating the tension to the centripetal force: $T = F_c$ $k(r - \ell_0) = 4\pi^2 f^2 m r$

(a) Radius of the uniform circular motion: Rearranging the equation to solve for $r$: $kr - k\ell_0 = 4\pi^2 f^2 m r$ $kr - 4\pi^2 f^2 m r = k\ell_0$ $r(k - 4\pi^2 f^2 m) = k\ell_0$ $r = \frac{k\ell_0}{k - 4\pi^2 f^2 m}$

(b) Tension $T$ in the cord: Using the relation $T = 4\pi^2 f^2 m r$ and substituting the expression for $r$: $T = 4\pi^2 f^2 m \left( \frac{k\ell_0}{k - 4\pi^2 f^2 m} \right)$ $T = \frac{4\pi^2 f^2 m k\ell_0}{k - 4\pi^2 f^2 m}$