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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 located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 0.52 rad/s2. What is the moment of inertia of the door about the hinges

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lublana

Answer:

[tex]7.69kgm^2[/tex]

Explanation:

We are given that

Mass,m=72 kg

Force,F=5 N

Distance,r=0.8 m

Angular acceleration,[tex]\alpha=0.52rad/s^2[/tex]

We have to find the moment of inertia of the door about hinges.

We know that

Torque,[tex]\tau=Fr=5\times 0.8=4Nm[/tex]

Moment of inertia,[tex]I=\frac{\tau}{\alpha}[/tex]

Using the formula

[tex]I=\frac{4}{0.52}=7.69Kgm^2[/tex]

Hence, the moment of inertia of the door about hinges=[tex]7.69kgm^2[/tex]

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