Answer to Question #179992 in Physics for ace

Question

Cord A is 6.00 m long, has breaking strength of 235. N, and is used to connect two balls of mass 500. g each. Cord B is identical to Cord A and has one end connected to one of the balls. The other end is held in order to swing the balls in a horizontal circle. As the balls are swung faster and faster, (a) which of the cords will break first? (b) At what angular speed, in revolutions per second, will the break occur?

Answer 1

Find the centripetal force acting on the ball that is further from the center (ball B). This force will be equal to the tension in cord B:

$$T=m \omega^2·2r,\\\space\\ \omega=\sqrt{\frac{T}{2mr}}.$$

By analogy, find the angular velocity for the ball closer to the center (ball A) from the tension in cord A. Remember that in this case, not only ball A creates the tension in cord A, but ball B as well:

$$T=m \omega^2·r+m\omega^22r, \\\space\\ \omega=\sqrt{\frac{T}{3mr}}.$$

Thus, cord A will break first.

Find the value of the angular velocity:

$$\omega=\sqrt{\frac{235}{3·0.5·6}}=5.1\text{ rad/s},\\\space\\ n=\frac{\omega}{2\pi}=0.81\text{ s}^{-1}.$$