For this part, you can either use ωf = ωi + αt to solve for t, then use the same old equation to solve for θ, or you can use ωf^2 - ωi^2 = 2αθ right off the bat.......
1) w_i = 0 rad/s
w_f = 53 rad/s
t = 10 s
Assume:
theta_i = 0 rad
Find the angular acceleration
alpha = (w_f - w_i) / t
Equation of motion
theta_f = theta_i + w_i*t + 1/2*alpha*t^2
Find theta_f, which I will call theta_1
theta_f = theta_1 = 0 + 0 + 1/2 * (w_f - w_i) * t
2) w_i = 53 rad/s
t = 40 s
Since there is no acceleration, alpha = 0. I'll call this part, theta_2
theta_2 = 0 + w_i*t + 0
3) w_i = 53 rad/s
w_f = 0 rad/s
alpha = -2.2 rad/s^2
alpha = (w_f - w_i) / t
t = (w_f - w_i ) / alpha
Again, assume:
theta_i = 0 rad
This part I'll call theta_3
theta_3 = 0 + w_i*t + 1/2*alpha*t^2 = w_i*t + 1/2*(w_f - w_i)^2 / alpha
Total amount of angular displacement = theta = theta_1 + theta_2 + theta_3
There are 2*pi radians per revolution
Number of revs = theta / (2*pi)
Separate the problem into three parts, solving for θ (the angle rotated through by the wheel) in each part, and add the results together at the end:
Part 1: the acceleration period
Givens:
ωi = 0rad/s
ωf = 53rad/s
t = 10s
α = ?
θ = ?
The equation we'll use to solve for θ is θ = (ωi)t + (1/2)αt^2. But first, we need to find α
ωf = ωi + αt
=> α = (ωf - ωi)/t
α = (53rad/s - 0rad/s)/10s
α = 5.3rad/s^2
Now, use the first equation to solve for θ during the acceleration period
θ = (ωi)t + (1/2)αt^2
θ = (0rad/s)(10s) + (1/2)(5.3rad/s^2)(10s)^2
θ = 265rad
Part 2: the constant velocity period
Givens:
ωi = 53rad/s
ωf = 53rad/s
α = 0rad/s^2
t = 40s
θ = ?
Use the same equation you used to solve for θ in the first part
θ = (ωi)t + (1/2)αt^2
θ = (53rad/s)(40s) + (1/2)(0rad/s^2)(40s)^2
θ = 2120rad
Part 3: the deceleration period
Givens:
ωi = 53rad/s
ωf = 0rad/s
α = -2.2rad/s^2
t = ?
θ = ?
For this part, you can either use ωf = ωi + αt to solve for t, then use the same old equation to solve for θ, or you can use ωf^2 - ωi^2 = 2αθ right off the bat.
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