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Sergio039
1 month ago
15

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 loc

ated 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?
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
3 0

To tackle this issue it is essential to use principles related to Torque and Inertia.

Torque is defined as the force exerted at a certain radius, meaning

\tau = F*r

Substituting our values we calculate

\tau = 5*0.8

\tau = 4N.m

On the other hand, Inertia is described as being inversely proportional to angular acceleration and directly proportional to torque, which means

I = \frac{\tau}{\alpha}

I = \frac{4}{2}

I = 2 kg.m^2

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