Question

A professor's office door is 0.99 m wide, 2.1 m high, 4.0 cm thick; has a mass of 26 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 5.5 N⋅m .
What is the moment of inertia of the door? If you let go of the open door, what is its angular acceleration immediately afterward?

Answer 1

Answer:

a. $I=8.4942 kg*m^2$

b. $a_c=0.647 m/s^2$

Explanation:

The fin the moment of inertia of the door:

Using the equation of the inertia in the pivots it means that use the equation of inertia in the the axis of rotation is the wide so:

a.

$I=\frac{1}{3}*m*r^2$

$I=\frac{1}{3}*26kg*(0.99m)^2$

$I=8.4942 kg*m^2$

b.

The torque of the door closer knowing the torque is the relation between the inertia and the angular acceleration

$T=I*a_c$

$a_c=\frac{T}{I}=\frac{5.5N*m}{8.4942kg*m^2}$

$a_c=0.647 m/s^2$