A 4.1-m-diameter merry-go-round is initially turning with a 4.2 period. It slows down and stops in 24 .
Before slowing, what is the speed of a child on the rim?
pi*4.1m/4.2u = 3.07m/u
To find the speed of a child on the rim of the merry-go-round, we can use the formula for linear speed:
Linear speed (v) = 2πr / T
Where:
- v is the linear speed
- r is the radius of the merry-go-round
- T is the period of rotation
Given that the diameter of the merry-go-round is 4.1 m, we can determine the radius (r) by dividing the diameter by 2:
r = 4.1 m / 2
r = 2.05 m
The period of rotation (T) is given as 4.2 seconds.
Now, we can substitute the values into the formula to calculate the linear speed:
v = 2π(2.05 m) / 4.2 s
v ≈ 3.069 m/s
Therefore, before slowing down, the child on the rim of the merry-go-round is moving at a speed of approximately 3.069 m/s.