An insect is at the bottom of a hemispherical | vedclass

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(D) Let $	heta$ be the angle the radius makes with the vertical at the point where the insect starts slipping. At this point,the component of gravity

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$0.20$

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(D) Let $	heta$ be the angle the radius makes with the vertical at the point where the insect starts slipping. At this point,the component of gravity along the tangent balances the maximum static friction.$mg \sin 	heta = f_{max} = \mu N$Since the normal force $N = mg \cos 	heta$,we have:$mg \sin 	heta = \mu mg \cos 	heta$$	an 	heta = \mu = 0.75 = \frac{3}{4}$From the geometry of the hemisphere,the height $h$ from the bottom is given by:$h = R - R \cos 	heta = R(1 - \cos 	heta)$Given $	an 	heta = \frac{3}{4}$,we have $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$.Substituting the values:$h = 1 	imes (1 - \frac{4}{5}) = 1 	imes \frac{1}{5} = 0.2\, m$.

An insect is at the bottom of a hemispherical ditch of radius $R = 1\, m$. 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 $\mu = 0.75$,then $h$ is $.......\, m$. $(g = 10\, m s^{-2})$

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