A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem.

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Question:

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem.

Part A
Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α.
Express your answer in terms of ω1 and α.
Part B
Find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1.
Express your answer in terms of some or all of the following: ω1, α, and t1.
Part C
Assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).
Express your answer in terms of some or all of the following: ω1, α, and t1.

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Answer:

(A) The time taken by the flywheel for the acceleration is, ?f=%5Cdfrac%7B%5Comega %7B1%7D%7D%7B%5Calpha%7D.

(B) The angle turned by the flywheel is ?f=%5Cdfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2 %7B1%7D.

(C)  The time taken by the flywheel to stop after the brake is applied is ?f=%5Cdfrac%7Bt %7B1%7D%7D%7B%205%7D.

Given data:

The magnitude of angular acceleration is, ?f=%5Calpha.

Final angular speed is, ?f=%5Comega %7B1%7D.

Time taken is, ?f=t %7B1%7D.

(A)

Apply the first rotational equation of motion as,

?f=%5Comega %7B1%7D%3D%5Comega'%20%2B%20%5Calpha%20t %7B1%7D

Since, the flywheel was initially at rest. So, ?f=%5Comega'%3D0.  Then,

?f=%5Comega %7B1%7D%3D0%20%2B%20%5Calpha%20%5Ctimes%20t %7B1%7D%5C%5Ct %7B1%7D%3D%5Cdfrac%7B%5Comega %7B1%7D%7D%7B%5Calpha%7D

Thus, the time taken for the acceleration is, ?f=%5Cdfrac%7B%5Comega %7B1%7D%7D%7B%5Calpha%7D.

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(B)

Now apply the second rotational equation of motion to obtain the angle turned by the flywheel as,

?f=%5Ctheta %7B1%7D%3D%5Comega%20'%20%5Ctimes%20t %7B1%7D%2B%5Cdfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2 %7B1%7D

Substituting the values as,

?f=%5Ctheta %7B1%7D%3D0%20%5Ctimes%20t %7B1%7D%2B%5Cdfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2 %7B1%7D%5C%5C%5Ctheta %7B1%7D%3D%5Cdfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2 %7B1%7D

Thus, the angle turned by the flywheel is ?f=%5Cdfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2 %7B1%7D.

(C) During acceleration, the expression is given as,

?f=%5Comega %7B1%7D%3D%5Comega'%20%2B%20%5Calpha%20t %7B1%7D%5C%5C%5Comega %7B1%7D%3D0%20%2B%20%5Calpha%20t %7B1%7D%5C%5Ct %7B1%7D%3D%5Cdfrac%7B%5Comega %7B1%7D%7D%7B%5Calpha%7D

Since, brake is applied. Which means the flywheel is stopped finally. So, applying the third rotational equation of motion as,

?f=%5Comega %7B2%7D%3D%5Comega %7B1%7D%2B( 5%20%5Calpha)t'

Here, ?f=%5Comega %7B2%7D is the final speed after brake is applied. And its value is ?f=%5Comega %7B2%7D%20%3D0.

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Solving as,

?f=0%3D%5Comega %7B1%7D 5%20%5Calpha%20t%5C%5Ct'%3D%5Cdfrac%7B%5Comega %7B1%7D%7D%7B5%20%5Calpha%20%7D%20%5C%5Ct'%3D%5Cdfrac%7B1%7D%7B%205%7D%20%5Ctimes%20%5Cdfrac%7B%5Comega %7B1%7D%7D%7B%20%5Calpha%20%7D%20%5C%5Ct'%3D%5Cdfrac%7Bt %7B1%7D%7D%7B%205%7D

Thus, the time taken by the flywheel to stop after the brake is applied is ?f=%5Cdfrac%7Bt %7B1%7D%7D%7B%205%7D.

Learn more about the rotational kinematics here:

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