Question

A fixed polygon has $n$ sides, each side of length $a$. A thread of length $na$ is fixed to one corner of the polygon and on the other side thread is attached to ball. Thread is made straight perpendicular to one of the sides $AB$. Ball is given a velocity $V$ perpendicular to the thread, parallel to the side $AB$ as shown. Then the time after which it will collide with the polygon is: ($AB$ is the side of the polygon and neglect gravity)

• A. $\frac{\pi a}{V}(n+1)$
• B. $\frac{a}{V}(n+1)$
• C. $\left(\frac{3\pi na + 2\pi a}{2V}\right)$
• D. $\left(\frac{3\pi n^2 a - 2\pi a}{2V}\right)$

Answer 1

Answer: (A)

$\frac{\pi a}{V}(n+1)

Answer 2

The problem describes the motion of a ball attached to a thread, unwrapping from a fixed regular polygon.

Understanding the motion:

The ball's path is an involute of the polygon. As the ball moves, the thread unwraps from the polygon in a series of circular arcs.

1. First Arc:

   • The thread is initially fixed at corner A and perpendicular to side AB. The length of the thread is $L_1 = na$.
   • The ball moves in a circular arc with radius $R_1 = na$, centered at A.
   • This continues until the thread becomes taut against the next vertex, B.
   • For a regular $n$-sided polygon, the exterior angle is $2\pi/n$. This is the angle by which the thread's direction changes as it wraps around each corner. Therefore, the angular extent of each circular arc is $\Delta\theta = 2\pi/n$.
   • The length of the first arc is $S_1 = R_1 \Delta\theta = na \left(\frac{2\pi}{n}\right) = 2\pi a$.
   • The time taken for the first arc is $t_1 = \frac{S_1}{V} = \frac{2\pi a}{V}$.

2. Subsequent Arcs:

   • After the first arc, a length $a$ of the thread has wrapped around side AB. The effective length of the thread remaining free to extend from vertex B is $L_2 = na - a = (n-1)a$.
   • The ball then moves in a second circular arc with radius $R_2 = (n-1)a$, centered at B.
   • The angular extent of this arc is also $\Delta\theta = 2\pi/n$.
   • The length of the second arc is $S_2 = R_2 \Delta\theta = (n-1)a \left(\frac{2\pi}{n}\right)$.
   • The time taken for the second arc is $t_2 = \frac{S_2}{V} = \frac{(n-1)a (2\pi/n)}{V}$.

This process continues. For each subsequent arc, the effective radius of rotation decreases by $a$. The radii of the successive arcs will be $na, (n-1)a, (n-2)a, \dots, a$. There will be $n$ such arcs in total.

Total Time:

The total time $T$ is the sum of the times taken for all $n$ arcs:

$T = t_1 + t_2 + \dots + t_n$

$T = \frac{1}{V} (S_1 + S_2 + \dots + S_n)$

$T = \frac{1}{V} \left[ na \left(\frac{2\pi}{n}\right) + (n-1)a \left(\frac{2\pi}{n}\right) + \dots + a \left(\frac{2\pi}{n}\right) \right]$

Factor out $\frac{a (2\pi/n)}{V}$:

$T = \frac{2\pi a}{nV} [n + (n-1) + (n-2) + \dots + 1]$

The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

So, $T = \frac{2\pi a}{nV} \left[ \frac{n(n+1)}{2} \right]$

$T = \frac{2\pi a}{V} \frac{(n+1)}{2}$

$T = \frac{\pi a}{V} (n+1)$