1.

A wheel having moment of inertia 2 kg m^(2) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheel's rotation in one minute would be

Answer»

`(2pi)/15Nm`
`(pi)/12Nm`
`(pi)/15Nm`
`pi/18Nm`

Solution :`omega_(f)=omega_(i)-alphatrArr0=omega_(i)-ALPHAT`
`thereforealpha=omega_(i)//t,` where `ALPHA` is RETARDATION.
The torque on the wheel is GIVEN by
`tau=Ialpha=(Iomega_(i))/t=(I.2piv)/t=(2xx2xxpixx60)/(60xx60)`
`rArr=pi/15` Nm
This is the torque REQUIRED to stop the wheel in 1 min. (or 60 sec.).


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