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8 Questions around this concept.
An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is $1 / 3$. If the line joining the centre of the hemispherical surface to the insect makes an angle $\theta$ with the vertical, the maximum possible value of $\theta$ is given by
If the coefficient of friction between on insect and bowl is $\mu$ and the radius of the bowl is r . Find the maximum height to which insect can crawl up in the bowl.
An insect is at the bottom of a hemisphere 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 $\mathrm{b} / \mathrm{w}$ the ground and insect is 0.75 . Then h is : $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
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A block of mass $m$ is placed on a surface having vertical cross section given by $y=x^2 / 4$. If coefficient of friction is 0.5 , the maximum height above the ground at which block can be placed without slipping is:
An inclined plane is bent in such a way that the vertical cross-section is given by $y=\frac{x^2}{4}$ where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $\mu=0.5$, the maximum height in cm at which a stationary block will not slip downard is $\qquad$ cm.
If the coefficient of friction between insect and bowl is $\mu$ and the radius of the bowl is $r=\sqrt{3} \mathrm{~m}$. If the maximum height to which the insect can crawl up in the bowl is $h=r-y_{\text {the }}$ the value of y is -$\left(\mu=\frac{1}{-3}\right)$
An insect trapped in circular groove. If it completes its 20 revolutions in 1 min with constant speed then frequency of insect is :
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Motion of an Insect in the Rough Bowl
As the insect crawls up, limiting friction friction force decreases and the component of weight along the suface (driving force) will decrease. Let's assume the insect can crawl upto height 'h' before it starts slipping. At that moment the frictional force will be limiting friction force as shown in the figure.
Let m=mass of the insect, r=radius of the bowl, μ= coefficient of friction for limiting condition at point A
$
\begin{aligned}
& \quad \mathrm{R}=\mathrm{mg} \cos \theta \ldots(\mathrm{i}) \\
& \text { Limiting friction }- \\
& \mathrm{f}_{\mathrm{l}}=\mu \mathrm{R}=\mu \mathrm{mg} \cos \theta \\
& \mathrm{f}_{\mathrm{l}}=\mathrm{mg} \sin \theta \ldots(\mathrm{ii}) \\
& \text { From equation (1) and (2)- } \\
& \mathrm{mg} \sin \theta=\mu \mathrm{mgcos} \theta \\
& \tan \theta=\mu \\
& \tan ^2 \theta=\mu^2 \\
& \frac{\left(\mathrm{r}^2-\mathrm{y}^2\right)}{\mathrm{y}^2}=\mu^2 \\
& \mathrm{y}=\frac{\mathrm{r}}{\sqrt{1+\mu^2}} \\
& \mathrm{r}-\mathrm{h}=\frac{\mathrm{r}}{\sqrt{1+\mu^2}} \\
& \mathrm{~h}=\mathrm{r}\left[1+\frac{1}{\sqrt{1+\mu^2}}\right]
\end{aligned}
$
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