Question

A particle is projected at an angle ${60}^{\circ }$ with the horizontal with a speed $10\text{m/sec}$. Find angular velocity about origin when particle touches $X-$ axis again.

• A. 1 rad/s
• B. 2 rad/s
• C. 3 rad/s
• D. 4 rad/s

Answer 1

The correct option is A 1 rad/s

$\omega =\frac{v}{R}=\frac{u\sin 60}{R}$
Here R is range of projectile which is acting as radius of curvature..
Range of projectile, $\left(R\right)=\frac{u^2\sin 2\theta }{g}R=\frac{u^2\sin 2\theta }{g}=\frac{100\times \frac{\sqrt{3}}{2}}{10}=5\sqrt{3}$
$\omega =\frac{10\times \frac{\sqrt{3}}{2}}{5\sqrt{3}}=1\text{rad/s}$