Question

A professor's office door is 0.94 m wide, 2.2 m high, 4.0 cm thick; has a mass of 30 kg , and pivots on frictionless hinges. A "door closer" is attached to door and the top of the door frame. When the door is open and at rest, the door closer exerts a torque of 5.8 N⋅m .

Answer 1

here

moment of inertia

I = ( 1 / 3 ) * m * r^2

I = (1/3) * 30 * 0.94^2

I = 8.836 kgm^2

then

angular acceleration is

alpha = torque / moment of inertia

alpha = 5.8 / 8.836

alpha = 0.656 rad/s^2