A wheel having moment of inertia 2 kg m−
2
about its vertical axis, rotates at the rate of 60 rev/min
about its axis. The torque which can stop the wheel’s rotation in 1 min would be :
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Answer:
t=π/15 Nm
Explanation:
Moment of inertia of wheel = I= 2kgm^(-2)
Angular velocity = ω=60 rpm
1 rpm = 2π/60 rad per sec
So
ω=60rpm = 2π rad per sec
Angular Acceleration = α=ω/t= 2π/60=π/30
t=Iα
=2kgm^(-2) ×π/30
=π/15 Nm
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