The flywheel of a steam engine runs with a constant angular speed of 116 rev/min
ID: 1271128 • Letter: T
Question
The flywheel of a steam engine runs with a constant angular speed of 116 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.0 h.
a.) What is the magnitude of the constant angular acceleration of the wheel?
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 44 cm from the axis of rotation when the flywheel is turning at 58 rev/min?
d.) What is the magnitude of the net linear acceleration of the particle in the above question?
Explanation / Answer
Initial angualar speed N1 = 116 rev/ minute
Final angular speed N2= 0
Time = 2 hours.
Angualar accleration = {N2-N1}/Time = -116 / 120= -0.97 rev/ minute. = -0.97/60 = 0.0162 rev/s^2
? = 2?*0.0162 rad/s^2
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Initial angualar speed N1 = 116 rev/ minute
Final angular speed N2= 0
Average angular speed = 116 /2 = Total no of revolutions /Time
Total no of revolutions = 116 /2 * time = [116 /2]* 120=6960 revoluttons.
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What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 44 cm from the axis of rotation when the flywheel is turning at 58 rev/min?
linear acceleration a = r ? = 0.44*2?*0.0162 = 0.0448 m/s^2
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What is the magnitude of the net linear acceleration of the particle in the above question?
Normal accelertion = r?^2
net acceleration =? [(r?^2)^2 + (r ?)^2] = r?[?^4 + ?^2]
? {0.44^2 *(2?*58/60} ^4 + (0.044) ^2 =16.23 m/s^2