A 5.0 m diameter merry-go-round is initially turning with a 4.6 s period. What is the speed of a child on the rim?
T = (2pir)/v with T being the period, r being the radius, and v being the speed. Don't forget that radius is half the diameter.
So, v*4.6s = 2*pi*2.5m
To find the speed of a child on the rim of the merry-go-round, we need to use the formula for linear velocity:
v = 2πr / T
where v is the linear velocity, π is a constant (approximately 3.14), r is the radius of the merry-go-round, and T is the period.
Given:
Diameter of the merry-go-round = 5.0 m
Radius (r) = Diameter / 2 = 5.0 m / 2 = 2.5 m
Period (T) = 4.6 s
Substituting the values into the formula:
v = (2 * 3.14 * 2.5) / 4.6
v ≈ 3.41 m/s
Therefore, the speed of a child on the rim of the merry-go-round is approximately 3.41 m/s.
To calculate the speed of a child on the rim of a merry-go-round, we need to use the formula for linear velocity.
The linear velocity of an object moving in a circle can be calculated using the formula:
v = 2πr / T
Where:
v is the linear velocity
π (pi) is a mathematical constant approximately equal to 3.14159
r is the radius of the circle
T is the period of the circular motion
In this case, we are given the diameter of the merry-go-round, which is 5.0 m. The radius is half of the diameter, so the radius would be 2.5 m. The period is given as 4.6 s.
Now we can plug these values into the formula to find the linear velocity of the child on the rim:
v = 2π(2.5) / 4.6
Calculating this, we get:
v ≈ 3.42 m/s
Therefore, the speed of the child on the rim of the merry-go-round is approximately 3.42 m/s.