A flywheel rotates with an angular speed of 26 rev/s. As it is brought to rest with a constant acceleration, it turns 55 rev.
What is the magnitude of the angular acceleration?
How much time does it take to stop?
To find the magnitude of the angular acceleration, we can use the equations of rotational motion. Let's denote the final angular speed as ωf, the initial angular speed as ωi, the angular acceleration as α, the total angle turned as θ, and the time taken as t.
Given:
ωi = 26 rev/s
θ = 55 rev
We know that the final angular speed is zero because the flywheel comes to rest.
Formula 1: ωf = ωi + αt
Formula 2: θ = ωit + (1/2)αt^2
Using Formula 1, we can solve for α:
ωf = 0 (since the flywheel comes to rest)
0 = 26 rev/s + αt
Solving for t, we have:
t = -26 rev/s / α (Equation 3)
Using Formula 2, we can solve for t as well:
θ = ωit + (1/2)αt^2
55 rev = (26 rev/s)t + (1/2)αt^2
Simplifying further:
(1/2)αt^2 + (26 rev/s)t - 55 rev = 0
Since we know that the flywheel comes to rest and θ = 55 rev, there are two solutions to this quadratic equation. The first solution will give negative values for t, so we discard that.
Using the quadratic formula:
t = (-26 rev/s ± √(26^2 rev^2/s^2 - 4(1/2)(-55 rev)(1))) / (2(1/2)α)
t = (-26 rev/s ± √(676 rev^2/s^2 + 55 rev)) / α
t = (-26 rev/s ± √(731 rev^2/s^2)) / α
t = (-26 rev/s ± (27 rev/s)) / α
t = (-53 rev/s or 1 rev/s) / α (Equation 4)
Now, equating the values of t from Equation 3 and Equation 4, we have:
-26 rev/s / α = -53 rev/s / α
α = -26 rev/s / -53 rev/s
α = 0.49 rev/s^2
Therefore, the magnitude of the angular acceleration is 0.49 rev/s^2.
To find the time it takes to stop, let's substitute the value of α back into Equation 3:
t = -26 rev/s / α
t = -26 rev/s / 0.49 rev/s^2
t = -52.98 s
Therefore, it takes approximately 53 seconds to bring the flywheel to rest.
To solve this problem, we can use the equations of rotational motion.
The first equation relates the angular displacement (θ) to the initial angular velocity (ω₀), angular acceleration (α), and time (t):
θ = ω₀t + (1/2)αt² [Equation 1]
The second equation relates the final angular velocity (ω) to the initial angular velocity, angular acceleration, and displacement:
ω = ω₀ + αt [Equation 2]
Given that the initial angular velocity (ω₀) is 26 rev/s and the final angular displacement (θ) is 55 revolutions, we want to find the magnitude of the angular acceleration (α) and the time it takes to stop (t).
First, let's find the magnitude of the angular acceleration (α):
Rearranging Equation 2, we have:
α = (ω - ω₀) / t
We know that the final angular velocity (ω) is 0 (because the flywheel comes to rest), so we can substitute that in:
α = (0 - 26 rev/s) / t
Simplifying, we get:
α = -26 rev/s / t
Now, let's find the time it takes to stop (t):
Using Equation 1, we can rearrange it to solve for t:
θ = ω₀t + (1/2)αt²
Since the flywheel comes to rest (ω = 0), we can substitute that in:
0 = ω₀t + (1/2)αt²
Now we can solve this quadratic equation for t. Rearranging, we get:
(1/2)αt² + ω₀t = 0
Plugging in the known values, we have:
(1/2)(-26 rev/s)t² + (26 rev/s)t = 0
Simplifying, we can cancel out common factors and solve for t:
-13t² + 26t = 0
Dividing both sides by t, we have:
-13t + 26 = 0
Now we can solve for t by dividing both sides by -13:
t = 26 / 13
t = 2 seconds
Therefore, the magnitude of the angular acceleration is -26 rev/s², and it takes 2 seconds for the flywheel to come to rest.
t = 55rev * (1/26)s/rev 2.1s.
a = (Vf - Vo) / t = (0 - 26) / 2.1 =
-12.4m/s^2.