A merry-go-round revolves at a rate of 8.3 rev/min. A child who is seated at the outer edge of the merry-go-round travels at 6.9m/s. What is the diameter of the merry-go-round?
v =ω•R = ω•D/2.
ω = 2•π•f ,
f = 8.3 rev/min = 8.3/60 =0.138 rev/s
D = 2•v/ ω = 2•v/ 2•π•f ω =
= 6.9/(3.14•0.138)=15.9 m
To find the diameter of the merry-go-round, we can use the relationship between linear speed and angular speed.
Given:
Rate of revolution of the merry-go-round: 8.3 rev/min
Linear speed of the child: 6.9 m/s
Step 1: Convert the rate of revolution to angular speed.
Angular speed (ω) = 2πf, where f is the frequency or rate of revolution in Hz.
Given that the rate of revolution is 8.3 rev/min, which is equivalent to 8.3/60 Hz (since 1 min = 60 s),
ω = 2π(8.3/60) Hz
ω = 0.27489 rad/s (approximately)
Step 2: Find the radius of the merry-go-round.
Linear speed (v) = ωr, where r is the radius.
Given that the linear speed of the child is 6.9 m/s and angular speed is 0.27489 rad/s,
6.9 = 0.27489r
r = 6.9 / 0.27489
r ≈ 25.11 m (approximately)
Step 3: Calculate the diameter of the merry-go-round.
Diameter (d) = 2r
d = 2(25.11)
d ≈ 50.22 m (approximately)
Therefore, the diameter of the merry-go-round is approximately 50.22 meters.
To find the diameter of the merry-go-round, we can use the relation between the linear speed of a point on the edge of a rotating object and its angular speed.
We know that the merry-go-round revolves at a rate of 8.3 rev/min, which means it completes 8.3 rotations every minute. To get the angular speed, we need to convert this to radians per second.
There are 2π radians in a full circle, so to convert revolutions per minute to radians per second, we can use the following conversion factor:
1 revolution/min = (2π radians)/(60 seconds)
Therefore, the angular speed (ω) of the merry-go-round is:
ω = (8.3 rev/min) * ((2π radians)/(60 seconds))
≈ 0.87267 radians/second
Now that we have the angular speed, we can use it to find the linear speed of a point on the edge of the merry-go-round. The linear speed (v) is related to the angular speed (ω) and the radius (r) of the merry-go-round by the equation:
v = ω * r
We know that the child seated on the outer edge of the merry-go-round travels at a linear speed of 6.9 m/s. Let's assume this distance is the radius (r) of the merry-go-round.
6.9 m/s = 0.87267 radians/second * r
Now we can solve for the radius (r):
r = (6.9 m/s) / (0.87267 radians/second)
≈ 7.9181 meters
Since the diameter of a circle is twice its radius, the diameter of the merry-go-round is:
Diameter = 2 * r
≈ 2 * 7.9181 meters
≈ 15.8362 meters
Therefore, the diameter of the merry-go-round is approximately 15.8362 meters.