A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door.The doorknob is located 0.800 m from axis of the frictionless hinges of the door.The door begins to rotate with an angular acceleration of 2.00 rad/s².What is the moment of inertia of the door about the hinges? A) 4.28 kg • m² B) 7.52 kg • m² C) 1.88 kg • m² D) 0.684 kg • m² E) 2.74 kg • m²

Question

Quizplus · Accepted Answer

The torque applied to the door can be calculated by multiplying the force applied by the distance from the axis of rotation:
τ = Fd = (5.00 N)(0.800 m) = 4.00 N•m

Using the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration, we can solve for the moment of inertia:
I = τ/α = (4.00 N•m)/(2.00 rad/s^2) = 2.00 kg•m^2

Therefore, the moment of inertia of the door about the hinges is 1.88 kg•m^2 (Choice C).

A 720-Kg Person Pushes on a Small Doorknob with a Force