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

A wheel of moment of inertia `2 kgm^(2)` is rotating about an axis pas

1.	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 isA. `24 kg m^(2)//s`B. `48 kg m^(2)//s`C. `72 kg m^(2)//s`D. `96 kg m^(2)//s`
Answer» Correct Answer - C 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`.