Answer to Question #127843 in Classical Mechanics for Shreyansh

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 μ=0.5. If half angle of cone is 45°, the maximum angular velocity of revolution of cone can be :

Answer 1

As per the question,

Height $(h)=1m$

Friction coefficient $(\mu) = 0.5$

Angle of cone $(\theta ) =45^\circ$

Let the maximum angular velocity is $\omega$

now,

Hence, radius $(r)$ of the rotating circle,

$\tan(45^\circ)=\frac{r}{h}$

$r=h\tan (45^\circ) = 1$

So, $N= mg\cos 45^\circ+ mr\omega^2 \cos 45^\circ$

$=\frac{mg}{\sqrt{2}}+\frac{mr\omega^2}{\sqrt{2}}$

$=\frac{mg+mr\omega^2}{\sqrt{2}}$

$=\frac{m(g+r\omega^2)}{\sqrt{2}}$

So, $f_s =\mu N = \frac{0.5\times m(g+r\omega^2)}{\sqrt{2}}=\frac{m(g+r\omega^2)}{2\sqrt{2}}$

Hence, $mg\sin 45^\circ =\frac{m(g+r\omega^2)}{2\sqrt{2}}$

$\Rightarrow 2g=g+ r\omega^2$

$\Rightarrow \omega^2=g$

$\Rightarrow \omega =\sqrt{10}=\sqrt{9.8} rad/sec$