Question

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:

The moment of inertia is $I = 1.8 \ kg m^2$

Explanation:

From the question we are told that

   The force applied is  $F = 10 \ N$

    The distance of the knob to the hinge is  $d = 0.9 \ m$

     The angular acceleration is  $a = 5 \ rad/s$

The moment is mathematically represented as

        $I = \frac{d Fsin(\theta)}{a}$

Here $\theta = 90^o$ This is because the force direction is perpendicular to the plane of the door

substituting values

          $I = \frac{0.9 * 10 * sin (90)}{5}$

          $I = 1.8 \ kg m^2$