Question

In a downtown office building, you notice each of the four sections of a rotating door has a mass of 75 kg. What is the width, in meters, of each section of the door if a force of 50 N applied to the outer edge of a section produces an angular acceleration of 0.410 rad/s2?

Answer 1

1.22 m

Step-by-step explanation:

Let the width of door is w.

mass, m = 75 kg

Force, F = 50 N

angular acceleration, α = 0.410 rad/s^2

Moment of inertia of the door, I = 4/3 mw^2

Torque = I x α = F x w

4/3 x 75 x w² x 0.410 = 50 x w

41 x w = 50

w = 1.22 m

Thus, the width of the door is 1.22 m.