Question

A child is sitting on the outer edge of a marry go round that is 18m in diameter. If the merry go round makes 8.3 rev/min what is the velocity of the child in m/s?

Answer 1

We know that

• The diameter is 18 meters long.

,

• The angular speed is 8.3 rev/min.

According to the definition of liner speed, we have the following formula.

$v=\omega r$

Where the radius is half the diameter: r = 18m/2 = 9m and the angular speed is 8.3 rev/min. But, first, we have to transform the angular speed into rad/s.

$8.3\cdot\frac{rev}{\min}\cdot\frac{2\pi\text{rad}}{1\text{rev}}\cdot\frac{1\min}{60\sec}=0.87\frac{rad}{\sec }$

Now we can find the linear speed.

$v=0.87\frac{rad}{\sec}\cdot9m=7.83\frac{m}{s}$

Therefore, the velocity of the child is 7.83 m/s.