Consider regular polygons with number of sides n = 3, 4, 5 ..... as shown in the figure. The center

Q: Why does the polygon roll about the vertex and not smoothly like a wheel?

A polygon has flat sides and sharp corners. When pushed, it pivots around the corner (vertex) in contact with the ground until the next flat side hits the ground. It cannot roll smoothly because its radius from the center to the perimeter is not constant.

Q: How did we get the angle $\pi/n$ in the triangle?

A regular polygon with $n$ sides can be divided into $n$ identical isosceles triangles from the center. The total angle at the center is $2\pi$, so each triangle has a central angle of $2\pi/n$. Dropping a perpendicular from the center to the base splits this angle in half, giving $\pi/n$.

Q: Why is the maximum height equal to $R$?

As the polygon pivots around the vertex $B$, the distance from the center $O$ to $B$ remains constant at $R$. The highest point of any circular path is when the radius vector is perfectly vertical. At this instant, the height from the ground is exactly the radius $R$.

Q: Does the mass of the polygon affect the answer?

No, this is a purely kinematic and geometric problem. We are only analyzing the trajectory of the center of mass based on the shape of the object, not the forces or energy required to make it roll.

Q: What happens if the polygon slips while rolling?

If the polygon slips, the pivot point $B$ would move horizontally during the rotation. The path of the center of mass would no longer be a simple circular arc, and the maximum height would depend on the complex dynamics of the slipping motion. The "no slipping" condition ensures a pure rotation about the vertex.

Question

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 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 :

Options:

(A) $\Delta = h \sin^2\left(\frac{\pi}{n}ight)$

(B) $\Delta = h \sin\left(\frac{2\pi}{n}ight)$

(C) $\Delta = h \left[ \frac{1}{\cos\left(\frac{\pi}{n}ight)} - 1 ight]$

(D) $\Delta = h 	an^2\left(\frac{\pi}{n}ight)$

Sigma Percentile · Accepted Answer

### The Bumpy Ride of a Polygon

Imagine trying to ride a bicycle with square wheels. It would be a jarring, bumpy experience! This happens because the center of mass of a square doesn't stay at a constant height when it rolls, unlike a perfectly round circular wheel.

In this problem, we are analyzing exactly how "bumpy" this ride is for any regular polygon with $n$ sides. We want to find the maximum increase in height, $\Delta$, of its center of mass as it rolls.

### Analyzing the Geometry

Let's start by looking at the polygon when it's resting flat on the ground. Its center of mass, $O$, is at an initial height $h$.

When the polygon begins to roll, it pivots around its leading vertex, which we will call $B$. To understand the motion, we need to find the distance from the center $O$ to this pivot vertex $B$. Let's call this distance $R$.

If we drop a perpendicular from the center $O$ to the ground, it hits the midpoint of the bottom side at point $A$. This creates a right-angled triangle, $	riangle OAB$.

In a regular polygon with $n$ sides, the total angle around the center is $2\pi$. This angle is divided equally among the $n$ sides, so each side subtends an angle of $\frac{2\pi}{n}$ at the center. The line $OA$ bisects this angle, so the angle $\angle AOB$ is exactly $\frac{\pi}{n}$.

### The Master Equation for Radius

Now, we can use simple trigonometry in $	riangle OAB$ to find $R$. The side adjacent to the angle $\frac{\pi}{n}$ is $OA$, which has a length of $h$. The hypotenuse is $OB$, which has a length of $R$.

Therefore, we can write:
$$ \cos\left(\frac{\pi}{n}ight) = \frac{h}{R} $$

Rearranging this equation gives us the crucial radius $R$:
$$ R = \frac{h}{\cos\left(\frac{\pi}{n}ight)} $$

### The Peak of the Roll

As the polygon rolls without slipping, it acts like a pendulum swinging upside down. The center of mass $O$ rotates in a perfect circular arc around the fixed pivot point $B$. The radius of this circular path is $R$.

The center of mass will reach its absolute highest point when it is directly vertically above the pivot vertex $B$. At this exact instant, the height of the center of mass from the ground is simply the radius $R$.

### Final Calculation

The question asks for the maximum increase in height, $\Delta$. This is simply the difference between the maximum height and the initial height.

$$ \Delta = R - h $$

Substituting our expression for $R$, we get:
$$ \Delta = \frac{h}{\cos\left(\frac{\pi}{n}ight)} - h $$

Factoring out the initial height $h$, we arrive at our final, elegant result:
$$ \Delta = h \left( \frac{1}{\cos\left(\frac{\pi}{n}ight)} - 1 ight) $$

**Sanity Check:** What happens if we increase the number of sides $n$ to infinity? The polygon transforms into a perfect circle. As $n 	o \infty$, the angle $\frac{\pi}{n} 	o 0$. Since $\cos(0) = 1$, the term inside the bracket becomes $1 - 1 = 0$. Thus, $\Delta = 0$. A perfect circle rolls smoothly without its center of mass bobbing up and down, which perfectly matches our physical intuition!

Select Answer:

Visualized Solution

$h_{initial} = h$

$∠ A O B = \frac{π}{n}$

$cos (\frac{π}{n}) = \frac{h}{R}$

$R = \frac{h}{c o s ( \frac{π}{n} )}$

$Path of COM$

$H_{max} = R$

$Δ = R - h$

$Δ = h (\frac{1}{c o s ( \frac{π}{n} )} - 1)$

$n \to \infty ⟹ Δ \to 0$

The Sigma Insight: Rolling Motion

The Bumpy Ride of a Polygon

Analyzing the Geometry

The Master Equation for Radius

The Peak of the Roll

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } hinitial​=h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∠AOB=nπ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(nπ​)=Rh​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=cos(nπ​)h​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path of COM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hmax​=R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ=R−h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ=h(cos(nπ​)1​−1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n→∞⟹Δ→0