JEE Main & Advanced Physics NLM, Friction, Circular Motion Motion of an Insect in the Rough Bowl

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 \[\cot \alpha =\frac{1}{\mu }=3\].