Question

Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round is 14 feet and it takes 6 seconds for it
to make a complete revolution? Round the solutions to two decimal places.

Answer 1

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s$

Also,

$v=r\omega$

Where

$\omega$ is the angular speed

So,

$\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s$

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.