Question

An ant starts crawling from the bottom of a bowl of radius $r$, having coefficient of friction $\mu$. The maximum vertical height $h$ up to which it can reach is

• A. $r[1+\frac{1}{\sqrt{1+{\mu }^2}}]$
• B. $r[1-\frac{1}{\sqrt{1+{\mu }^2}}]$
• C. $r[1+\frac{1}{\sqrt{1-{\mu }^2}}]$
• D. $r[1-\frac{1}{\sqrt{1-{\mu }^2}}]$

Answer 1

The correct option is B $r[1-\frac{1}{\sqrt{1+{\mu }^2}}]$

At the point of the maximum vertical height, let the radius make angle $\theta$ with the vertical.

Then, the FBD of the ant can be drawn.


Along the radial direction,
$N=mgcos\theta$

Along the tangential direction,
$f=mgsin\theta$

In the limiting condition (ant is just about to slip),
$f=\mu N=\mu mgcos\theta =mgsin\theta$

$\Rightarrow \mu =tan\theta$

From the geometry of $\Delta OAB$,
$tan\theta =\frac{y}{\sqrt{r^2-y^2}}=\mu$

$\Rightarrow y=\mu \sqrt{r^2-y^2}$

$\Rightarrow y^2={\mu }^2(r^2-y^2$

$\Rightarrow y=\frac{r}{\sqrt{1+{\mu }^2}}$

Therefore, the maximum vertical distance,
$h=r-y=r-\frac{r}{\sqrt{1+{\mu }^2}}$

$=r(1-\frac{1}{\sqrt{1+{\mu }^2}})$