A Ferris wheel with a diameter of 35.0 m in starts from rest and achieves its maximum operational tangential speed of 2.20 m/s in a time of 15.0 s. (a) What is the magnitude of the wheel's angular acceleration? (b) What is the magnitude of the tangential acceleration after the maximum operational speed is reached?
To find the magnitude of the wheel's angular acceleration, we can use the formula:
Angular acceleration (α) = Change in angular velocity (Δω) / Time (t)
Since the wheel starts from rest, the initial angular velocity (ωi) is 0. The final angular velocity (ωf) can be found using the formula:
ωf = 2π * f
where f is the frequency of rotation. Since we are given the tangential speed (v) instead of the frequency, we can use the formula:
v = ω * r
where r is the radius of the wheel.
Given that the diameter of the wheel (d) is 35.0 m, the radius (r) can be found by dividing the diameter by 2:
r = d / 2 = 35.0 m / 2 = 17.5 m
Substituting the given tangential speed (v = 2.20 m/s) and the radius (r = 17.5 m) into the formula for angular velocity, we have:
2.20 m/s = ω * 17.5 m
ω = 2.20 m/s / 17.5 m
ω = 0.1257 rad/s
The change in angular velocity (Δω) between 0 rad/s and 0.1257 rad/s is:
Δω = ωf - ωi = 0.1257 rad/s - 0 rad/s = 0.1257 rad/s
Finally, substituting this change in angular velocity and the given time (t = 15.0 s) into the formula for angular acceleration, we get:
Angular acceleration (α) = Δω / t
α = 0.1257 rad/s / 15.0 s
α ≈ 0.00838 rad/s²
Therefore, the magnitude of the wheel's angular acceleration is approximately 0.00838 rad/s².
To find the magnitude of the tangential acceleration after the maximum operational speed is reached, we can use the formula:
Tangential acceleration (at) = α * r
Substituting the calculated angular acceleration (α ≈ 0.00838 rad/s²) and the radius (r = 17.5 m) into the formula, we get:
Tangential acceleration (at) = 0.00838 rad/s² * 17.5 m
at ≈ 0.146 m/s²
Therefore, the magnitude of the tangential acceleration after the maximum operational speed is reached is approximately 0.146 m/s².
To find the magnitude of the wheel's angular acceleration, we can use the formula:
Angular acceleration (α) = (final angular velocity - initial angular velocity) / time
The diameter of the Ferris wheel is given as 35.0 m, which means the radius (r) is half of that value:
r = 35.0 m / 2 = 17.5 m
The initial angular velocity (ω₀) is 0 because the wheel starts from rest.
The final angular velocity (ω) can be calculated using the formula:
ω = tangential velocity / radius
The tangential velocity (v) when the wheel achieves its maximum operational speed is given as 2.20 m/s.
ω = 2.20 m/s / 17.5 m
ω = 0.1257 rad/s
Using the information given, we can now calculate the angular acceleration:
α = (ω - ω₀) / time
α = (0.1257 rad/s - 0 rad/s) / 15.0 s
α = 0.00838 rad/s²
Therefore, the magnitude of the wheel's angular acceleration is 0.00838 rad/s².
To find the magnitude of the tangential acceleration after the maximum operational speed is reached, we can use the formula:
Tangential acceleration (aₜ) = angular acceleration (α) * radius
Using the previously calculated value of α and the radius (r) of 17.5 m:
aₜ = 0.00838 rad/s² * 17.5 m
aₜ = 0.146 rad/s²
Therefore, the magnitude of the tangential acceleration after the maximum operational speed is reached is 0.146 m/s².
2.2 m/s is an angular speed of 2.2/35π = 0.02 rad/s
so, its angular acceleration is
(.02rad/s)/(15s) = 0.00133 rad/s^2
I expect you can find the other one.