Question

A child is sitting on the outer edge of a merry-go-round that is 1.5 m in diameter. If the merry-go-round makes 3.2 rev/min, what is the velocity of the child in m/s?

Answer 1

Answer:

the velocity of the kid is 5.6 m/s

Explanation:

r is the radius and w is the frequency.

so we should know that the diameter is 18m and the diameter is equal to two times the radius, so r = 18m/2 = 9m

we should also know that the circumference of a circle is equal to c = 2pi*r, so each revolution has this length. if the kid does 5.9 revolutions in one minute then the kid spins at v = 5.9*2pi*9m/min

so we want to write this in meters per second and this means that we need to divide it by 60!

v = (5.9*2pi*9/60)m/s = 5.56 m/s

so your answer will be  5.6 m/s glad i could help!

Answer 2

0.25 m/s

Explanation:

The radius r of the merry-go-round is half its diameter D:

$r = \frac{1}{2}D = \frac{1}{2}(1.5\:\text{m}) = 0.75\:\text{m}$

We also need to convert the angular speed from rev/min to rad/s. We know that there are 60 seconds to a minute and that there are $2\pi$ radians per revolution. Therefore,

$\omega = 3.2\:\dfrac{\text{rev}}{\text{min}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}}×\dfrac{1\:\text{min}}{60\:\text{s}}$

$\:\:\:\:=0.335\:\text{rad/s}$

Now that we know the angular speed in rad/s, the child's linear speed can be calculated as

$v = r\omega = (0.75\:\text{m})(0.335\:\text{rad/s}) = 0.25\:\text{m/s}$