Consider regular polygons with number of sides n = 3 , 4 , 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

Question

Consider regular polygons with number of sides $n = 3, 4, 5, \dots$ 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 $Δ$ . Then $Δ$ depends on $n$ and $h$ as [2017]

Smart Achievers · Accepted Answer

(2)

When the polygon rolls without slipping about the point $' P'$ . The point O reaches the maximum height (Point O' in the figure) by moving in a circle of radius $' r'$ about the point 'P'.

So, maximum increase in height of centre of mass 'O' is given by $Δ = r - h = \frac{h}{\cos (\frac{π}{n})} - h$

$\Rightarrow Δ = h (\frac{1}{\cos (\frac{π}{n})} - 1)$

Solution

1 $Δ = h \sin^{2} (\frac{π}{n})$

2 $Δ = h (\frac{1}{\cos (\frac{π}{n})} - 1)$

3 $Δ = h \sin (\frac{2 π}{n})$

4 $Δ = h \tan^{2} (\frac{π}{2 n})$

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Solution

1 Δ=hsin2(πn)

2 Δ=h(1cos(πn)-1)

3 Δ=hsin(2πn)

4 Δ=htan2(π2n)

Ans. (2) When the polygon rolls without slipping about the point 'P'. The point O reaches the maximum height (Point O' in the figure) by moving in a circle of radius 'r' about the point 'P'. So, maximum increase in height of centre of mass 'O' is given by Δ=r-h=hcos(πn)-h ⇒Δ=h(1cos(πn)-1)

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1 $Δ = h \sin^{2} (\frac{π}{n})$

2 $Δ = h (\frac{1}{\cos (\frac{π}{n})} - 1)$

3 $Δ = h \sin (\frac{2 π}{n})$

4 $Δ = h \tan^{2} (\frac{π}{2 n})$