Question

A particle inside the rough surface of a rotating cone about its axis is at rest, relative to it as a height of 1m above its vertex. Friction coefficient is $\mu =0.5$. If half angle of cone is ${45}^0$, the maximum angular velocity of revolution of cone can be:

• A. $\sqrt{10}$ rad/s
• B. $\sqrt{30}$ rad/s
• C. $\frac{\sqrt{40}}{3}$ rad/s
• D. $\sqrt{50}$ rad/s

Answer 1

The correct option is A $\sqrt{10}$ rad/s

$Given$ :- Mass of particle be ( $m$ )

Half angle of cone = 45°

Coefficient of friction ( $\mu$ ) = 0.5

Acceleration due to gravity $\left(g\right)$ = 10 ${m/s}^2$

Height of particle from vertex $\left(r\right)$ = 1m

$ToFind$ :- Maximum angular velocity ( $\omega$ )

$Solution$ :- we know , for max $\omega$ , friction $\left(f\right)$ = 0

Now since the particle is at rest ,

$\therefore$ $mgsin45°=F_{Centripetal}sin45°$

$⟹$ $mg=F_{Centripetal}$

$mg=mr{\omega }^2$

$⟹$ $\omega =\sqrt{\frac{g}{r}}$

$\therefore$ $\underset{–}{\omega =\sqrt{10}rad/s}$

Hence , $\underset{–}{OptionA\left(\sqrt{10}rad/s\right)iscorrect.}$