Question

A force of 860 N is applied at a point 1.9 m from the axis of rotation, causing a revolving door to accelerate at 426.52 deg/s2. What is the moment of inertia of the door from its axis of rotation

Answer 1

Answer:

3.629 Kg $m^{2}$

Explanation:

Force=F= 860 N

Distance from Center of rotation= r= 1.9 m

Angular acceleration= α = 426.52 deg/sec2

We use the relation;

τ = I α

where,

τ = Torque

I = moment of inertia

==> F× r = I ×426.52

==> 860×1.8 = I × 426.52

==> 1548 =  I × 426.52

==> I = $\frac{1548}{426.52}$ =  3.629 Kg $m^{2}$