Aforce of 300 n is applied at a point 1.3 m from the axis of rotation, causing a revolving door to accelerate at 309.4 deg/s2. what is the moment of inertia of the door from its axis of rotation?

Moment of inertia,                                                                                  

Explanation:

It is given that,

Force, F = 300 N

Distance from the axis of rotation, r = 1.3 m

Acceleration,

In linear kinematics, torque is given by :

.............(1)

In rotational kinematics, torque is given by :

.........(2)

I is the moment of inertia

From equation (1) and (2) :

So, the moment of inertia of the door from its axis of rotation is . Hence, this is the required solution.

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