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(rev)/(\min)\cdot\frac{2\pi\text{rad}}{1\text{rev}}\cdot(1\min)/(60\sec)=0.87(rad)/(\sec )`

Now we can find the linear speed.

`v=0.87(rad)/(\sec)\cdot9m=7.83(m)/(s)`

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