A 5.2-m-diameter merry-go-round is initially turning with a 3.0s period. It slows down and stops in 28s.
i found the v = 5.4 m/s, How many revolutions does the merry-go-round make as it stops?
Well, if the merry-go-round stops, that means it's going from going round and round to being stuck in one place. So, the number of revolutions it makes as it stops is exactly zero! It went from spinning to not spinning at all. Talk about going from hero to zero!
To find the number of revolutions the merry-go-round makes as it stops, we need to calculate the initial angular velocity and final angular velocity.
The formula for angular velocity (ω) is given as:
ω = 2π / T
where ω is the angular velocity and T is the period.
Given that the initial period (T₁) is 3.0 s, we can find the initial angular velocity (ω₁):
ω₁ = 2π / T₁
= 2π / 3.0
≈ 2.09 rad/s
Next, we need to find the final angular velocity (ω₂). The final angular velocity is zero because the merry-go-round stops. We can calculate the angular acceleration (α) using the formula:
α = (ω₂ - ω₁) / t
where α is the angular acceleration, ω₁ is the initial angular velocity, ω₂ is the final angular velocity, and t is the time taken for the merry-go-round to stop.
Given that t = 28 s, we can calculate α:
0 = (ω₂ - 2.09) / 28
Solving for ω₂, we get:
ω₂ = 2.09 rad/s
Now, we can calculate the number of revolutions (N) using the formula:
N = (θ₂ - θ₁) / 2π
where N is the number of revolutions, θ₁ is the initial angle, and θ₂ is the final angle.
Since the merry-go-round initially has an angular displacement of 0 degrees, θ₁ = 0.
To find θ₂, we can use the kinematic equation:
ω₂² = ω₁² + 2αθ
Substituting the values, we have:
(2.09)² = (2.09)² + 2(2.09)θ
Simplifying this equation and solving for θ, we get:
θ = -0.996 rad
Now, we can calculate the number of revolutions:
N = (-0.996 - 0) / (2π)
≈ -0.159
Since we're interested in the absolute value of the number of revolutions, we round it to:
N ≈ 0.159
Therefore, the merry-go-round makes approximately 0.159 revolutions as it stops.
To calculate the number of revolutions the merry-go-round makes as it stops, we need to find the initial and final angular velocities and then use the equation relating angular velocity, time, and the number of revolutions.
Given:
Diameter of the merry-go-round, d = 5.2 m
Initial period, T_initial = 3.0 s
Time it takes to stop, t_final = 28 s
First, let's find the initial angular velocity (ω_initial) using the formula:
ω_initial = 2π / T_initial
where ω is the angular velocity and T is the period.
Substituting the given values, we have:
ω_initial = 2π / 3.0 s ≈ 2.094 radians/s
Next, let's find the final angular velocity (ω_final) using the formula:
ω_final = 0 (since the merry-go-round stops, the final angular velocity is 0)
Now, we can use the equation that relates angular velocity, time, and the number of revolutions:
θ = ω_initial * t_final
where θ is the angle covered in radians.
Substituting the values, we have:
θ = 2.094 radians/s * 28 s ≈ 58.632 radians
To find the number of revolutions (N), we need to convert this angle to revolutions. Since there are 2π radians in one revolution, we can use the formula:
N = θ / (2π)
Substituting the value of θ, we get:
N = 58.632 radians / (2π) ≈ 9.35 revolutions
Therefore, the merry-go-round makes approximately 9.35 revolutions as it stops.
Average rev/min = (1/2)*(20 rev/min)
= 10 rev/min
No. of revolutions = 10 rev/min*(28/60 min = 4.66
You do not need to know the diameter.