A $14.5 ext{ kg}$ mass, fastened to the end of a steel wire of unstretched length $1.0 ext{ m}$, i

Question

A $14.5	ext{ kg}$ mass, fastened to the end of a steel wire of unstretched length $1.0	ext{ m}$, is whirled in a vertical circle with an angular velocity of $2	ext{ rev/s}$ at the bottom of the circle. What is the approximate tension in the wire when the mass is at the lowest point of its path? (Take $g = 9.8	ext{ m/s}^2$ and $\pi^2 \approx 9.87$)

Professor PC Thomas Classes · Answer

At the bottom, Tension $T = mg + m\omega^2 L$. Angular velocity $\omega = 2	ext{ rev/s} = 4\pi	ext{ rad/s}$. $\omega^2 = 16\pi^2 \approx 157.9$. $T = 14.5(9.8 + 157.9 	imes 1.0) = 14.5(167.7) \approx 2432	ext{ N}$. Using $\pi^2 \approx 10$, $\omega^2 = 160$, $T = 14.5(169.8) \approx 2462	ext{ N}$.