Question

A small sized mass $m$ is attached by a massless string (of length $L$) to the top of a fixed frictionless solid cone whose axis is vertical. The half angle at the vertex of the cone is $\theta$. If the mass $m$ moves around in a horizontal circle at speed $v$, what is the maximum value of $v$ for which the mass stays in contact with the cone? ($g$ is acceleration due to gravity)

• A. $\sqrt{gL\cos \theta }$
• B. $\sqrt{gL\sin \theta }$
• C. $\sqrt{gL\sin \theta \tan \theta }$
• D. $\sqrt{gL\tan \theta }$

Answer 1

The correct option is C $\sqrt{gL\sin \theta \tan \theta }$

From FBD of mass:


$N$ is the normal reaction on the mass from the solid cone. When the mass is on the verge of losing contact, $N=0$

$T\cos \theta$ balances the weight of the bob and $T\sin \theta$ provides centripetal force.
$T\cos \theta =mg$ ... (1)
$T\sin \theta =mr{\omega }^2$ ... (2)
From (1) and (2)
$\tan \theta =\frac{r{\omega }^2}{g}$
$\Rightarrow {\omega }^2=\frac{g\tan \theta }{r}$
$\Rightarrow \omega =\sqrt{\frac{g\tan \theta }{r}}$
From the question diagram, $r=L\sin \theta$
To find $v$:
We know that $v=r\omega$
$\therefore v=r\sqrt{\frac{g\tan \theta }{r}}$
$=\sqrt{rg\tan \theta }$
$\therefore v=\sqrt{gL\sin \theta \tan \theta }$