A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0 kg and
diameter 80.0 cm. The power is off for 32.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes180 complete revolutions.
A. At what rate is the flywheel spinning when the power comes back on? (rad/s)
B. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on? (s)
C. How many revolutions would the wheel have made during this time?
A. At what rate is the flywheel spinning when the power comes back on? (rad/s)
B. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on? (s)
C. How many revolutions would the wheel have made during this time?
1 answer:
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Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 =
ω2 = 216.47
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