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A flywheel of moment of inertia 250 kg m^(2) is rotating at an angular speed of 12 rad s^(-1). What torque is needed to stop the wheel in 6 s ? |
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Answer» Solution :`omega_(2)=12" rad/s, "omega=0,t=6s` `THEREFORE alpha=(omega-omega_(0))/(t)=(0-12)/(6)=-2" rad/"s^(2)` `therefore` retardation = `2" rad/"s^(2)` `therefore` RETARDING TORQUE, `tau=Ialpha=250xx2=500Nm` |
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