Motion of an Insect in the Rough Bowl
Category : JEE Main & Advanced
Motion of an Insect in the Rough Bowl The insect crawl up the bowl up to a certain height h only till the component of its weight along the bowl is balanced by limiting frictional force. Let m = mass of the insect, r = radius of the bowl, m = coefficient of friction For limiting condition at point A \[R=mg\cos \theta \] ...... (i) and \[{{F}_{l}}=mg\sin \theta \] ...... (ii) Dividing (ii) by (i) \[\tan \theta =\frac{{{F}_{l}}}{R}=\mu \] \[\left[ \text{As}\,\,{{F}_{l}}=\mu R \right]\] \[\therefore \] \[\frac{\sqrt{{{r}^{2}}-{{y}^{2}}}}{y}=\mu \] or \[y=\frac{r}{\sqrt{1+{{\mu }^{2}}}}\] So \[h=r-y=r\,\left[ 1-\frac{1}{\sqrt{1+{{\mu }^{2}}}} \right]\], \[\therefore \,\,\,\,h=r\,\left[ 1-\frac{1}{\sqrt{1+{{\mu }^{2}}}} \right]\] Problem 22. 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 \[\alpha \] with the vertical, the maximum possible value of \[\alpha \] is given by [IIT-JEE (Screening) 2001] (a) \[\cot \alpha =3\] (b) \[\tan \alpha =3\] (c) \[\sec \alpha =3\] (d) \[\text{cosec}\,\alpha =\text{3}\] Solution: (a) From the above expression, for the equilibrium \[R=mg\,\cos \,\alpha \,\text{and}\,F=mg\,\sin \,\alpha \]. Substituting these value in \[F=\mu R\] we get \[\tan \alpha =\mu \] or \[\cot \alpha =\frac{1}{\mu }=3\].
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