Answer:
The period of motion is 1.26x10⁻⁶ s.
Explanation:
The period of motion can be found as follows:
[tex] \omega = \frac{2 \pi}{T} [/tex]
Where:
ω is the angular speed
T is the period
The angular speed is related to tangential speed (v):
[tex] v = \omega r [/tex]
r is the radius
Hence, the period is:
[tex] T = \frac{2 \pi r}{v} = \frac{2 \pi 0.2 m}{10^{6} m/s} = 1.26 \cdot 10^{-6} s [/tex]
Therefore, the period of motion is 1.26x10⁻⁶ s.
I hope it helps you!
