Question

A pencil of hexagonal cross-section of side $a$ is placed on a rough surface as shown.

The platform starts performing S.H.M. perpendicular to length of the pencil in horizontal plane with angular frequency $\omega$. Considering there is sufficient friction between pencil and platform such that there is no slipping between them, the maximum amplitude of oscillation such the pencil does not topple is $\frac{g}{\sqrt{\alpha }{\omega }^2}.$ Then the value of $\alpha$ is (Answer upto two digits after the decimal point)

Answer 1

The acceleration of the platform is $A{\omega }^2$, where $A$ is its amplitude.

Taking frame of reference as the platform, there will be a pseudo force of $m{\omega }^{2}A$ on the block acting at its COM as shown in the figure.
By balancing the forces, we get,
$f=m{\omega }^{2}A;N=mg$
By balancing the torque about COM, we get,
$f\frac{\sqrt{3}a}{2}=N\frac{a}{2}$
$\frac{m{\omega }^{2}Aa\sqrt{3}}{2}=mg\frac{a}{2}A=\frac{g}{\sqrt{3}{\omega }^2}$
$\alpha =3$