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7 June, 23:04

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a constant tangential force of 260 n applied to its edge causes the wheel to have an angular acceleration of 0.810 rad/s2. (a) what is the moment of inertia of the wheel?

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  1. 8 June, 01:27
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    Refer to the diagram shown below.

    Let I = the moment of inertia of the wheel.

    α = 0.81 rad/s², the angular acceleration

    r = 0.33 m, the radius of the weel

    F = 260 N, the applied tangential force

    The applied torque is

    T = F*r

    = (260 N) * (0.33 m)

    = 85.8 N-m

    By definition,

    T = I*α

    Therefore,

    I = T/α

    = (85.8 N-m) / (0.81 rad/s²)

    = 105.93 kg-m²

    Answer: 105.93 kg-m²
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