Question

An insect is crawling in a hemi-spherical bowl of radius \( R \). If the coefficient of friction between the insect and bowl is \( \mu \), then the maximum height to which the insect can crawl the bowl is

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

Hint:

When solving problems involving motion on curved surfaces with friction, always decompose forces into tangential and normal components to find the limiting conditions for motion.

Answer 1

The Correct Option is A

To find the maximum height \( h \) that the insect can reach, we consider the balance of forces acting on the insect. The key forces are the gravitational force and the frictional force that prevents the insect from sliding down. At the maximum height, the normal force \( N \) provided by the surface of the bowl acts perpendicular to the surface, and the gravitational force \( mg \) acts downwards. The frictional force \( f \) that prevents the insect from sliding must satisfy the relation: \[ f = \mu N \] At the point where the insect stops climbing, the tangential component of gravitational force (which tries to slide the insect down) equals the frictional force. If \( \theta \) is the angle from the vertical at this point, then: \[ mg \sin \theta = \mu mg \cos \theta \] \[ \tan \theta = \mu \] Using the geometry of the hemisphere, the height \( h \) can be related to \( \theta \) by: \[ h = R - R \cos \theta \] \[ \cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} \] \[ \cos \theta = \frac{1}{\sqrt{1 + \mu^2}} \] Thus, the maximum height \( h \) is: \[ h = R - R \frac{1}{\sqrt{1 + \mu^2}} \] \[ h = R \left[1 - \frac{1}{\sqrt{1 + \mu^2}}\right] \]