A door of moment of inertia 4 km `m^(2)` is at rest .When a torque of 2`pi` Nm acts on it find its angular acceleration Find also its angular velocity after 1S

Explanation

Moment of inertia I=4kg `m^(2) omega_(1)=0,tau=2piNm`
i)Angular acceleration `alpha=?,tau=Ialpha=(tau)/(I)=(2pi)/(4)=(pi)/(2) rad//s^(2)`
`omega_(1)=o,omega_(2)=?t=1 sec,omega -omega_(1)=alphat,omega_(2)=(pi)/(2)xx(1),omega_(2)=(3.14)/(2)=1.57 `rad/sec