Question

A vertical hollow cylinder is fixed on the ground. A uniform rod can be balanced partly in and partly out of the cylinder with the lower end of the rod resting against the vertical wall of the cylinder, as shown in the figure. The angle made by rod with the vertical in equilibrium is $\theta$. Maximum and minimum possible value of $\theta$ are 53° and 37° respectively. Coefficient of friction between rod and cylinder is $\mu =\tan \left[\frac{1}{2}{\tan }^{-1}\left(\frac{37}{14\eta }\right)\right]$. The value of $\eta$ is. (Answer upto two digits after the decimal point)

Answer 1

Suppose the radius of the cylinder is $R$ and length of rod is $2l$.
Consider the case when the end a has sliding tendency upwards.
Forces acting on the rod are shown in the figure.

Resolving forces horizontally and vertically and then balancing them.
We have
$\begin{array}{}{N}_2=N_1\cos \alpha +\mu N_1\sin \alpha & \text{(1)}\end{array}$
and $\begin{array}{}{N}_1\sin \alpha =\mu N_2+\mu N_1\cos \alpha +W& \text{(2)}\end{array}$
Taking moment about A.
$\begin{array}{}{N}_1\left(2R\text{cosec}\alpha \right)=W\left(l\sin \alpha \right)& \text{(3)}\end{array}$
From equation (1),(2) and (3), we get
$\begin{array}{}2R=l\left[\right(1-{\mu }^2)\sin \alpha -2\mu \cos \alpha ]{\sin }^2\alpha & \text{(4)}\end{array}$

Similarly, when the rod makes least angle $\beta$, we have sliding tendency downward.

$\begin{array}{}2R=l\left[\right(1-{\mu }^2)\sin \beta +2\mu \cos \beta ]{\sin }^2\beta & \text{(5)}\end{array}$
From equation (4) and (5), we get

$(1-{\mu }^2)(si{n}^3\alpha -si{n}^3\beta )=2\mu ({\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta )$

$⟹\frac{(si{n}^3\alpha -si{n}^3\beta )}{({\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta )}=\frac{2\mu }{1-{\mu }^2}$

Take $\mu =tan\left(\theta \right)$

$⟹\frac{(si{n}^3\alpha -si{n}^3\beta )}{({\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta )}=\frac{2tan\left(\theta \right)}{1-ta{n}^2\left(\theta \right)}$

$=\frac{(si{n}^3\alpha -si{n}^3\beta )}{({\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta )}=tan\left(2\theta \right)$

$⟹\theta =\frac{1}{2}ta{n}^{-1}\left(\frac{(si{n}^3\alpha -si{n}^3\beta )}{({\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta )}\right)$
so
$\mu =\tan \left[\frac{1}{2}{\tan }^{-1}\left(\frac{{\sin }^3\alpha -{\sin }^3\beta }{{\sin }^2\alpha \cos \alpha +{\sin }^2\beta \cos \beta }\right)\right]$
taking the values of $\alpha \text{and}\beta$ we get

$\mu =\tan \left[\frac{1}{2}{\tan }^{-1}\left(\frac{37}{84}\right)\right]$

Hence $\eta =6$