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An electrical motor is rotating at 900rpm and when it is swirched off it is brought to rest with uniform retardation in 400 revolutions . Find the time taken to bring the motor rest and the angular retardation .

Gracious Mambwe , 6 Years ago
Grade 12th pass
anser 1 Answers
Karan Gupta

Last Activity: 6 Years ago

Initial angular velocity = 900 rpm = 900/60 rev/s = 15 rev/s = 2π*(15) = (30π) rad/s
Final angular velocity = 0
Angular displacement = 400 rev = 2π*(400) = (800π) rad
ω2 = ωo2 + 2αθ
0 = (30π)^2 – 2α (800π)
α = 9π/16 rad/s2
 
From ω = ωo +  αt
we get, 0 = 30π – (9π/16)t
 
Means, t = (160/3) seconds

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