Question

1. a track coach starts an analogue stopwatch and allows it to run for one minute. if the sweep second hand on the stopwatch is 2.0 cm in length, determine

a) the speed of the sweep second hand. (remember that the circumference of a circle is $c = 2\pi r$)
did you get an answer of $2.1\times10^{-3}\frac{m}{s}$?
b) the velocity of the second hand at the 15 s point. which way would it point?

Explanation:

Step1: Calculate the distance traveled by the second - hand

The second - hand of a stopwatch makes one full revolution in 60 s. The radius of the circle traced by the tip of the second - hand is $r = 2.0\ cm=0.02\ m$. The circumference of the circle is $C = 2\pi r$. So, $C = 2\pi\times0.02\ m$.

Step2: Calculate the speed

Speed $v=\frac{d}{t}$. Here, $d = C$ (distance traveled in one minute which is one full - circle) and $t = 60\ s$. So, $v=\frac{2\pi\times0.02}{60}\ m/s\approx2.1\times 10^{-3}\ m/s$.

Step3: Determine the velocity at 15 s

At 15 s, the second - hand is at the 3 o'clock position. The velocity of an object moving in a circle is tangent to the circle at the point of the object's position. So, the direction of the velocity of the second - hand at 15 s is perpendicular to the second - hand and points in the counter - clockwise (assuming the normal rotation of a clock) direction. The magnitude of the velocity is the same as the speed calculated above, $v = 2.1\times 10^{-3}\ m/s$.

Answer:

a) The speed of the sweep second hand is approximately $2.1\times 10^{-3}\ m/s$.
b) The magnitude of the velocity of the second hand at the 15 s point is $2.1\times 10^{-3}\ m/s$ and it points perpendicular to the second - hand in the counter - clockwise direction.