Question

A force 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?

Answer 1

Moment of inertia,
`I=1.26\ kg-m^2`

Step-by-step explanation:

It is given that,

Force, F = 300 N

Distance from the axis of rotation, r = 1.3 m

Acceleration,
`\alpha=309.4\ rad/s^2`

In linear kinematics, torque is given by :

`\tau=F* r`.............(1)

In rotational kinematics, torque is given by :

`\tau=I\alpha`.........(2)

I is the moment of inertia

From equation (1) and (2) :

`Fr=I\alpha`

`I=(Fr)/(\alpha )`

`I=(300* 1.3)/(309.4)`

`I=1.26\ kg-m^2`

So, the moment of inertia of the door from its axis of rotation is
`1.26\ kg-m^2`. Hence, this is the required solution.