Question

The flywheel of a steam engine runs with a constant angular speed of 121 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.1 h.

(A) What is the magnitude of the constant angular acceleration of the wheel in rev/min^2?

(B) How many rotations does the wheel make before coming to rest?

(C) What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 43 cm from the axis of rotation when the flywheel is turning at 60.5 rev/min?

(D) What is the magnitude of the net linear acceleration of the particle in the above question?

Answer 1

Initial angualar speed N1 = 121 rev/ minute
Final angular speed N2= 0
Time = 2.1 hours =126 min
Angualar accleration = {N2-N1}/Time = -121 / 126= -0.96 rev/ minute. = -0.96/60 = 0.016 rev/s^2
α = 2π*0.016 = 0.10048 rad/s^2

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Initial angualar speed N1 = 121 rev/ minute
Final angular speed N2= 0

Average angular speed = 121/2 = Total no of revolutions /Time
Total no of revolutions = 121/2 * time = [121 /2]* 126=7623 revoluttons.

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linear acceleration a = r α = 0.43*0.10048 = 0.043 m/s^2

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Normal accelertion = rω^2
net acceleration =√ [(rω^2)^2 + (r α)^2] = r√[ω^4 + α^2]

√ {0.43^2 *(2π*60.5/60} ^4 + (0.043) ^2
√ 0.1849*1606.29+0.001849=10.59 m/s^2