Question

A right circular cone with semi-vertical angle $\alpha$ rests on a rough incline plane. As angle of inclination $\theta$ is increased, the cone will slide before it topples over, if coefficient of friction :-

• A. $\mu <\tan \alpha$
• B. $\mu <\frac{3}{4}\tan \alpha$
• C. $\mu <4\tan \alpha$
• D. $\mu <\frac{4}{3}tan\alpha$

Answer 1

The correct option is C $\mu <4\tan \alpha$

Condition for sliding: $mg\sin \theta -f>0$, where $f$ is the limiting friction.

$\Rightarrow mg\sin \theta >\mu \left(mg\cos \theta \right)$

$\Rightarrow \mu <\tan \theta$

If the cone doesn't topple, then torque ($\tau$) $=0$ about the bottom-most point.

$\Rightarrow {\tau }_{{}_N}+{\tau }_{{}_f}+{\tau }_{{}_{mg}}=0$

We know that $mg$ acts on the center of mass, which for a cone is $\frac{h}{4}$ above the base.

Breaking $mg$ into $x$ and $y$ components, and using right-hand convention:

$\Rightarrow \left(0\right)+\left(0\right)+(r\times mg\cos \theta )-(\frac{h}{4}\times mg\sin \theta )=0$, where $r$ and $h$ are radius of the base and the height respectively.

$\Rightarrow \tan \theta =4\times \frac{r}{h}$

We know from earlier that $\mu <\tan \theta$

$\therefore \mu <4\times \frac{r}{h}$

Via geometry, we have $\tan \alpha =r/h$

$\therefore \mu <4\tan \alpha$

$\to QED.$