Question

The given figure shows the cross section of a uniform pencil placed on a rough platform. The cross-section of the pencil is a hexagon of side $‘a^{\prime }$. The platform starts performing S.H.M. perpendicular to the length of the pencil in horizontal plane with angular frequency $‘{\omega }^{\prime }$. There is sufficient friction between the pencil and the platform such that there is no slipping between them. The maximum amplitude of oscillations so that the pencil does not topple is $\frac{g}{\sqrt{\alpha }{\omega }^2}$. Find $\alpha$

Answer 1

From the FBD of the pencil:
$N=mg$
& $f=m{\omega }^{2}x$
where $x$ is the displacement from mean position.

So, maximum value of friction will be at extreme positions ($x=A$)
$f=m{\omega }^{2}A$

Taking torques about COM of pencil, condition for no toppling is
$f\times \frac{\sqrt{3}a}{2}\le N\times \frac{a}{2}$
$\Rightarrow m{\omega }^{2}A\sqrt{3}\le N=mg$
$\Rightarrow A\le \frac{g}{\sqrt{3}{\omega }^{{}^2}}$
Hence, $\alpha =3$