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

Answer:

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

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 :

$\tau=F\times r$.............(1)

In rotational kinematics, torque is given by :

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

I is the moment of inertia

From equation (1) and (2) :

$Fr=I\alpha$

$I=\dfrac{Fr}{\alpha }$

$I=\dfrac{300\times 1.3}{309.4}$

$I=1.26\ kg-m^2$

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.