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

Question

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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Diễm Kiều 14 hours 2021-07-21T18:13:02+00:00 1 Answers 0 views 0

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    2021-07-21T18:14:36+00:00

    Answer:

    7.69kgm^2

    Explanation:

    We are given that

    Mass,m=72 kg

    Force,F=5 N

    Distance,r=0.8 m

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

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

    We know that

    Torque,\tau=Fr=5\times 0.8=4Nm

    Moment of inertia,I=\frac{\tau}{\alpha}

    Using the formula

    I=\frac{4}{0.52}=7.69Kgm^2

    Hence, the moment of inertia of the door about hinges=7.69kgm^2

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