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A disk with a rotational inertia of 5kg*m^2 and a radius of 0.25 meters rotates on...

A disk with a rotational inertia of 5kg*m^2 and a radius of 0.25 meters rotates on a friction-less fixed axis perpendicular to the disk and through its center. A force of 2 Newtons is applied tangentially to the rim. After five seconds this force is removed and a braking force is applied. What amount of (tangential) braking force would be necessary to slow it down to a stop in two revolutions?

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Answer #1

I= IMR? - 1 * 5x (0-257²8.15625 kg ma Ta Fr=2x0.25=0.5 Id =P d = 1 = 0.5 3-2 rad/ 0.15625 W = Wotat = 0+3.26 5 16 rad/s. o=24

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