How far does the tip of a $5 \text{ foot}$ pendulum travel as it swings through an angle of $30^\circ$?

Question

How far does the tip of a $5 \text{ foot}$ pendulum travel as it swings through an angle of $30^\circ$?

I proceeded by drawing a picture of swinging pendulum with dimensions. I drew a right triangle next to pendulum with the length of pendulum $(5 \text{ foot})$ being the $x$-axis of my right triangle. $30^\circ (\theta)$ is the tangent angle. I solved for tangent and got $2.89$ for $y$-axis. I used the Pythagoras Theorem [Baudhāyana Śulbasûtra] to solve for length of the hypotenuse which would be my answer $C=5.77 \text{ foot}$.

I ask the community if my steps are sound in judgement or not?

Answer 1

You want the arc length of an arc spanning $30°$ with radius $r= 5.0$

1. Convert the angle into radians
2. Use the arc length formula $s = r\, \theta$

Answer 2

We have that the lenght of the pendulum is constant therefore the distance in horizontal direction is

$$d=H \cdot \sin 30° = 5 \cdot \frac12 = 2.50$$

and the arc length

$$s=R\cdot \theta=5\cdot \frac{30\pi}{180}\approx 2.62$$

Answer 3

Thirty degrees is one-twelfth of a complete circle (since a complete circle is $360$ degrees). The circumference of a circle of radius $R$ is $2\pi R$. So if a $5$-foot pendulum swings through an angle of $30$ degrees, its tip travels an arc-length distance of $(2\pi\times5)/12=5\pi/6\approx2.618$ feet, which can be rounded up to $2.62$ feet.

If by "distance traveled" you mean the linear distance between the two endpoints of the tip's $30$-degree swing, then the answer is

$$5\sqrt{2-2\cos(30^\circ)}=5\sqrt{2-\sqrt3}\approx2.58819\approx2.59\text{ feet}$$