Question

One end of a string of length L and cross-sectional area A is fixed to a support and the other end is fixed to a bob of mass m. The bob is revolved in a horizontal circle of radius r with an angular velocity $\omega$ such that the string makes an angle $\theta$ with the vertical. The angular velocity $\omega$ is equal to :

• A. $\sqrt{\frac{gsin\theta }{r}}$
• B. $\sqrt{\frac{gcos\theta }{r}}$
• C. $\sqrt{\frac{gtan\theta }{r}}$
• D. $\sqrt{\frac{gcot\theta }{r}}$

Answer 1

Answer: (C)

$\sqrt{\frac{gtan\theta }{r}}$

Draw the FBD of the BOB
Balancing the vertical forces we will get
$Tcos\theta =mg$
Balancing the horizontal forces we will get
$Tsin\theta =m{\omega }^{2}r$
dividing the above two equation we will get
$tan\theta =\frac{r{\omega }^2}{g}$
or $\omega =\sqrt{\frac{gtan\theta }{r}}$