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Suppose that a person closes a car door. The person's fingers are in the door opening, that is located 0.50 m from the hinges. The rotational inertia of the car door is 5.0 kg⋅m2. The door closes in 0.1 s and the rotational speed of the door is 1.4 rad/s just before it closes
Determine the average force that the car door exerts on the person's fingers

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  1. yep, it will hurt.

    torque*time=I*w
    force*distane*time=I*w
    you have distance, time, I, and w
    solve for force.
    I get on the order of 140 N

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