A door of moment of inertia 4 kg 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 1 s.

A door of moment of inertia 4 kg m^(2) is at rest. When a torque of 2

1.	A door of moment of inertia 4 kg 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 1 s.
Answer» Solution :`alpha=(TAU)/(I)=(2pi)/(4)=(PI)/(2)rad//s^(2)` `omega_(2)-omega_(1)=alphat""omega_(2)=(pi)/(2)XX1` `omega_(2)=1.57` rad/sec.