A 2.0 m diameter merry-go-round is initially turning with a 4.2 s period. It slows and stops in 22.0 seconds. Before slowing, what is the speed (in m/s) of a child on the rim?
How many revolutions does the merry-go-round make as it stops?
one rev in 4.2 seconds
v= 2PI*radius/4.2
figure that out
Now, average angular velocity= Vi/(2*r)
revolutions= avgangularveloicty*22sec
This is wrong
To find the speed of a child on the rim of the merry-go-round before it slows down, we need to use the formula for speed, which is given by:
Speed = Distance / Time
In this case, the distance traveled by the child on the rim of the merry-go-round is equal to the circumference of the merry-go-round, which is given by:
Circumference = Pi * Diameter
Substituting the values, we have:
Circumference = 3.14 * 2.0 m = 6.28 m
Next, we need to determine the time it takes for the merry-go-round to complete one revolution. This is called the period, which is given as 4.2 seconds.
Now, we can find the speed:
Speed = Circumference / Time
Speed = 6.28 m / 4.2 s
Simplifying the expression gives:
Speed = 1.496 m/s
Therefore, the speed of the child on the rim of the merry-go-round before it slows down is approximately 1.496 m/s.
To find the number of revolutions the merry-go-round makes as it stops, we need to determine the change in time. The initial period was 4.2 seconds and it stops after 22.0 seconds, so the change in time is:
Change in time = Final time - Initial time
Change in time = 22.0 s - 4.2 s
Change in time = 17.8 s
To find the number of revolutions, we divide the change in time by the initial period since each period represents one revolution:
Number of revolutions = Change in time / Initial period
Number of revolutions = 17.8 s / 4.2 s
Simplifying the expression gives:
Number of revolutions ≈ 4.24
Therefore, the merry-go-round makes approximately 4.24 revolutions as it stops.