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1 day ago

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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?

1 answer:

Yuliya22 [2.4K]1 day 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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