Question

A person pushes on a doorknob with a force of 5.00 N. The direction of the force is at an angle of 30.0° from the perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the hinges of the door. The door begins to rotate from rest with uniform angular acceleration and makes half a revolution in 2.23 seconds. What is the moment of inertia of the door about the hinges?

Answer 1

Number of revolutions made by the door in 2.23 sec = n = 0.5 rev

Time period = T = 2.23 sec

Angle through which the door rotates =

= 2n

= 2(0.5)

= 3.141 rad

Initial angular velocity of the door = = 0 rad/s (At rest)

Angular acceleration of the door =

= T + T²/2

3.141 = (0)(2.23) + (2.23)²/2

= 1.263 rad/s²

Force applied on the doorknob = F = 5 N

Angle the force makes with the perpendicular = = 30^o

Component of force perpendicular to the door = FCos

Distance of the doorknob from the hinges of the door = R = 0.8 m

Moment of inertia of the door about the hinges = I

I = FCosR

I(1.263) = (5)(0.8)Cos(30)

I = 2.742 kg.m²

Moment of inertia of the door about the hinges = 2.742 kg.m²