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

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

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

Step-by-step 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=(1)/(3)*m*r^2`

`I=(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=(T)/(I)=(5.5N*m)/(8.4942kg*m^2)`

`a_c=0.647 m/s^2`