A wheel of moment of inertia `2 kgm^(2)` is rotating about an axis passing through centre and perpendicular to its plane at a speed `60 rad//s`. Due to friction, it comes to rest in 5 minutes. The angular momentum of the wheel three minutes before it stops rotating is

Options

A. `24 kg m^(2)//s`
B. `48 kg m^(2)//s`
C. `72 kg m^(2)//s`
D. `96 kg m^(2)//s`

Answer

C

Explanation

Given. `I = 2 kg m^(2)`
`omega_(0) = 60` rad/s , `omega = 0`
`t = 5` min ` = 5 xx 60 = 300` s
From the relation,
`omega = omega_(0) + alphat`
`alpha = (omega - omega_(0))/(t)`
`rArr alpha = (0 -60)/(300) = (-60)/(300) = (-1)/(5) rad//s^(2)`
For `t = 3=2` min
`= 2 xx 60 = 120s`
`omega = omega_(0) + alphat`
`= 60 - 1/(5) xx 120 = 60 - 24`
`:. omega = 36` rad/s
As, angular momentum, `L =komega`
Substituting the values in the above relation,
we get `L = 2 xx 36 = 72 kg m^(2)//s`.