Question

g: To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door

Answer 1

Answer:

I =1.8 kgm^2

Explanation:

In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:

$\tau=I\alpha$        (1)

τ: torque applied to the door

I: moment of inertia of the door

α: angular acceleration = 5 rad/s^2

The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).

F = 10 N

d = 0.9 m

You replace the expression for τ, and solve for I:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2$

The moment of inertia of the door is 1.8 kgm^2