A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting from rest, gaining angular speed at the constant rate α1 through the first 400 rad and then losing angular speed at the constant rate -α1 until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 400 m/s2. (a) What is the least time required for the rotation? (b) What is the corresponding value of α1?
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To find the least time required for the rotation and the corresponding value of α1, we can break down the motion of the disk into two parts: the first 400 rad where it is gaining angular speed, and the second 400 rad where it is losing angular speed.
(a) To find the least time required for the rotation, we need to determine the time it takes for the disk to go through the first 400 rad and the time it takes for the disk to go through the second 400 rad.
For the first 400 rad:
We need to find the angular acceleration (α1) that ensures the magnitude of the centripetal acceleration does not exceed 400 m/s^2. The centripetal acceleration (a_c) is related to angular acceleration (α) and radius (r) by the equation:
a_c = rα
In this case, a_c is given as 400 m/s^2 and r is 0.25 m. Plugging these values into the equation, we have:
400 = 0.25α1
Solving for α1, we find:
α1 = 400 / 0.25
α1 = 1600 rad/s^2
Using the kinematic equation for angular motion:
θ = ω_0t + (1/2)αt^2
Where θ is the angular displacement, ω_0 is the initial angular velocity (which is 0 since the disk starts from rest), α is the angular acceleration (α1), and t is the time.
Plugging in the known values, we have:
400 = 0 + (1/2)(1600)t^2
Simplifying and solving for t, we find:
t^2 = 400 / 800
t = √(0.5)
t = 0.707 seconds
So, the time required for the first 400 rad is approximately 0.707 seconds.
For the second 400 rad:
The angular acceleration is now -α1 (negative value) because the disk is losing angular speed. Using the same kinematic equation as above, we have:
400 = 0 + (1/2)(-1600)t^2
Simplifying and solving for t, we find:
t^2 = -400 / -800
t = √(0.5)
t = 0.707 seconds
So, the time required for the second 400 rad is also approximately 0.707 seconds.
The total time required for the rotation is the sum of the times for the first and second 400 rad:
Total time = 0.707 + 0.707
Total time = 1.414 seconds
Therefore, the least time required for the rotation is approximately 1.414 seconds.
(b) The value of α1 is determined from the first part of the calculation:
α1 = 1600 rad/s^2
So, the corresponding value of α1 is 1600 rad/s^2.