The flywheel of an engine has a moment of inertia 3.00 kg m^2 about its rotation...
The flywheel of an engine has a moment of inertia 3.00 kg m^2
about its rotation axis.
What constant torque is required to bring it up to an angular speed
of 370 rev/min in a time of 7.60 s, starting from rest?
N?m
What is its final kinetic energy?
J
Homework Answers
here we have given with values
n = 370 rpm = 370/60 = 6.1667 rev/sec
? = 2?n =2 * 3.14 * 6.1667=38.72
we know that
L = I? = 3 * 38.72 = 116.18 N-m-s
we know that
? = ?L/?t = 116.18/ 7.60=15.2868 N-m
part b ) kinetic energy = Iw^2/2 = 0.5 * 3 * (38.72 ) ^2 = 2248.8576
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