An insect is crawling in a hemispherical bowl | vedclass

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Question

(A) At the maximum height,the insect is at the point of slipping. The forces acting on the insect are its weight $mg$ (downwards),the normal reaction

Vedclass · Accepted Answer

$R\left[1-\frac{1}{\sqrt{1+\mu^2}}\right]$

Vedclass · Answer

(A) At the maximum height,the insect is at the point of slipping. The forces acting on the insect are its weight $mg$ (downwards),the normal reaction $N$ (radially outwards),and the limiting friction $f = \mu N$ (tangentially upwards).Resolving the weight $mg$ into components,we have $mg \cos 	heta$ along the normal and $mg \sin 	heta$ along the tangent.For equilibrium in the normal direction: $N = mg \cos 	heta$.For equilibrium in the tangential direction: $f = mg \sin 	heta$.Since $f = \mu N$,we have $\mu N = mg \sin 	heta$.Substituting $N = mg \cos 	heta$,we get $\mu (mg \cos 	heta) = mg \sin 	heta$,which simplifies to $	an 	heta = \mu$.The height $H$ of the insect from the bottom of the bowl is given by $H = R - R \cos 	heta = R(1 - \cos 	heta)$.Using the identity $\cos 	heta = \frac{1}{\sec 	heta} = \frac{1}{\sqrt{1 + 	an^2 	heta}}$,and substituting $	an 	heta = \mu$,we get $\cos 	heta = \frac{1}{\sqrt{1 + \mu^2}}$.Therefore,the maximum height is $H = R\left(1 - \frac{1}{\sqrt{1 + \mu^2}}ight)$.

An insect is crawling in a hemispherical bowl of radius $R$. If the coefficient of friction between the insect and the bowl is $\mu$,then the maximum height to which the insect can crawl in the bowl is

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