Question

If the coefficient of friction between an insect and bowl is $\mu$ and the radius of the bowl is $R$. Then the maximum height to which the insect can crawl in the bowl is

• A. $\frac{R}{\sqrt{1+{\mu }^2}}$
• B. $R(1-\frac{1}{\sqrt{1+{\mu }^2}})$
• C. $R\sqrt{1+{\mu }^2}$
• D. $R\sqrt{1+{\mu }^2}-1$

Answer 1

The correct option is B $R(1-\frac{1}{\sqrt{1+{\mu }^2}})$

Assumptions:
(1) Bowl is hemispherical in shape.
(2) Insect is moving so slowly that we can assume it is in equilibrium at each point.
At highest point insect tries to slip down. so the friction force ($f$) acts tangentially upward.

F.B.D of insect at highest point is


Because insect is at equilibrium at any point, we can equate tangential and normal component of force.

In normal direction,
$N=mgcos\theta$
In tangential direction,
$f=mgsin\theta$

For limiting case of static friction,
$f=\mu N$ $⟹mgsin\theta =\mu mgcos\theta$

Hence, $\mu =tan\theta$


From diagram we can say the maximum height reached by the insect is
$h=R-Rcos\theta$
$\because tan\theta =\mu$


Hence, $cos\theta =\frac{1}{\sqrt{{\mu }^2+1}}$
So the maximum height reached by insect is,
$h=R(1-cos\theta )=R(1-\frac{1}{\sqrt{{\mu }^2+1}})$