Question

Consider regular polygons with number of sides $n=3,4,\mathrm{5.......}$ as shown in the figure. The center of mass of all the polygons is at height $h$ from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for each polygon is $\Delta$. Then $\Delta$ depends on $n$ and $h$ as

• A. $\Delta =h\sin \left(\frac{2\pi }{n}\right)$
• B. $\Delta =h{\sin }^2\left(\frac{\pi }{n}\right)$
• C. $\Delta =h{\tan }^2\left(\frac{\pi }{2n}\right)$
• D. $\Delta =h(\frac{1}{\cos \left(\frac{\pi }{n}\right)}-1)$

Answer 1

The correct option is D

$\Delta =h(\frac{1}{\cos \left(\frac{\pi }{n}\right)}-1)$

Let $n$ be the number of sides of a regular polygon . By symmmetry, its centre of mass $O$ will be at same distance from each vertex.
Let $r$ be the radius of circumscribed circle and $h$ be the perpendicular distance of $O$ from any side. The angle subtended by any side on the centre
$O$ is $\frac{2\pi }{n}$ and so half of the angle subtended $\angle POQ=\frac{\pi }{n}$



Here when the polygon rolls without slipping or sliding about the point $P$, the point $O$ moves with the circle of radius $r$ about the point $P$. The point $O$ reaches the maximum height
$($point $O^{\prime }$ in the figure$)$ when $P{O}^{\prime }$ is perpendicular to $PQ$. Thus, the maximum increase in height of the locus of the centre of mass $O$ is given by
$\Delta =r-h=\frac{h}{cos\left(\frac{\pi }{n}\right)}-h=h(\frac{1}{\cos \left(\frac{\pi }{n}\right)}-1)$