A fly-wheel of mass 25 kg has a radius of 0.2 m. It is making 240 rpm. What is the torque necessary to bring to rest in 20 s?

Options

A. `2piNM`
B. `0.4piNM`
C. `(2)/(pi)NM`
D. `4piNM`

Answer

(b)

Explanation

For a fly-wheel,
Mass, m= 25 kg
  Radius, r=0.2 m
Angular velocity, `omega = 240` rpm
Time, t= 20 s
From the torque equation,
`tau=Ia`
Where,
`a=(2pin)/(t)=(2pixx(240)/(60))/(20)=(2pixx4)/(20)=((2pi)/(5))`
`tau= MR^(2)a`
=`(25)(0.04)((2pi)/(5))=0.4pi Nm`.