Question

An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is :
(g = 10 \(\mathrm{ms^{-2}}\))

• A. 0.45 m
• B. 0.60 m
• C. 0.20 m
• D. 0.80 m

Hint:

At the point of slipping the insect is at an angle \(\theta\) from the vertical with \(\tan\theta = \mu\); then \(h = R(1-\cos\theta)\).

Answer 1

The Correct Option is C

Let the radius to the insect make angle \(\theta\) with the vertical. Forces: weight mg, normal N (along the radius) and friction f (along the tangent).
Along the tangent: \(f = mg\sin\theta\). Along the radius: \(N = mg\cos\theta\).
Slipping begins when \(f = \mu N\): \(\tan\theta = \mu = 0.75 = \frac34\), so \(\cos\theta = 0.8\).
Height from the bottom: \(h = R - R\cos\theta = 1 - 0.8 = 0.2\) m.
Answer: (3) 0.20 m