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A fly wheel of moment of inertia 4xx10^(...

A fly wheel of moment of inertia `4xx10^(-3)kgm^(2)` is makign 10 revolution per second. The torque required to stop it in 5 second is:

A

`4pixx10^(-3)Nm`

B

`2pixx10^(-4)Nm`

C

`8pixx10^(-4)NM`

D

`16pixx10^(-3)Nm`

Text Solution

Verified by Experts

The correct Answer is:
D
`omega=omega_(0)+alphat`
If `omega=0` then `0=omega_(0)+alphat`
Angular acceleration `alpha=(-omega_(0))/t`
Angular retardation `alpha=(omega_(0))/t`
`omega_(0)=2pin=2pixx10=20pi`
`t=5` second
`:.alpha=(20pi)/5`
`=4pi` rad/s
Torque required to stop the wheel
`tau=I alpha`
`tau=4xx10^(-3)xx4pi`
`=16pixx10^(-3)Nm`
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