Q. A wheel making 20 revolution per second is in a horizontal circle with a uniform angular velocity. Let $T$ be the tension in the string. If the length of the string is halved and its angular velocity is doubled, tension in the string will be
A) $\pi$ rad $s^{- 2}$
B) 2$\ \pi$ rad $s^{- 2}$
C) 4 $\pi$ rad $s^{- 2}$
D) 8 $\pi$ rad $s^{- 2}$
Explanation
(b) $\omega_{1} = 2\pi r = 2\pi \times 20,\omega_{2} = 0,t = 20a,\alpha = ?$
As $\ \omega_{2} = \omega_{1} + \ \alpha t$
$\therefore\ \alpha = \frac{\omega_{2} - \omega_{1}}{t} = \frac{40\pi - 0}{20} = 2\pi$ rad $s^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.