Question:
A bead of mass $m$ stays at point $P(a, b)$ on a wire bent in the shape of a parabola $y=4 C x^{2}$ and rotating with angular speed $\omega$ (see figure). The value of $\omega$ is (neglect friction) :
Correct Option: 1
Solution:
(1) $y=4 C x^{2} \Rightarrow \frac{d y}{d x}=\tan \theta=8 C x$
At $P, \tan \theta=8 \mathrm{Ca}$
For steady circular motion
$m \omega^{2} a \cos \theta=m g \sin \theta$
$\Rightarrow \omega=\sqrt{\frac{g \tan \theta}{a}}$
$\therefore \omega=\sqrt{\frac{g \times 8 a C}{a}}=2 \sqrt{2 g C}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.