1.

A wheel of moment of inertia 5xx10^(-3)kgm^(2) is making 20 rps. It is stopped in 20 s. The angular retardation is :

Answer»

`pi` rad `s^(-2)`
`2pi` rad `s^(-2)`
`4pi` rad `s^(-2)`
`8pi` rad `s^(-2)`

SOLUTION :`omega_(0)=2piv=40pis^(-1).omega=0`
`ALPHA=(omega-omega_(0))/(t)=(0-40pi)/(20)=-2pi" rad/"s^(2)`
`therefore` angular retardation= `2pi" rad/"s^(2)`


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