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CyclicCircle

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A constant tangential force of magnitude 12N is applied to the rim of a stationary, uniform circular flywheel of mass 100kg and radius 0.5m. Find the speed at which the flywheel is rotating after it has completed 25 revolutions?

I know that this can be done using work-energy. But since a constant tangential force is applied, I tried using kinematic equations.

Initial angular velocity [itex]\omega = 0[/itex], angular displacement [itex]\theta = 25 \times 2\pi = 50\pi[/itex].

If [itex]\alpha[/itex] is the angular acceleration, [itex]mr\alpha = 12[/itex], [itex](100)(0.5)\alpha = 12[/itex], [itex]\alpha = 0.24[/itex].

Final velocity [itex]\Omega^2 = \omega^2 + 2\alpha \theta[/itex], which gives [itex]\Omega = 8.68[/itex]. But the correct answer is apparently 12.3.

I know that this can be done using work-energy. But since a constant tangential force is applied, I tried using kinematic equations.

Initial angular velocity [itex]\omega = 0[/itex], angular displacement [itex]\theta = 25 \times 2\pi = 50\pi[/itex].

If [itex]\alpha[/itex] is the angular acceleration, [itex]mr\alpha = 12[/itex], [itex](100)(0.5)\alpha = 12[/itex], [itex]\alpha = 0.24[/itex].

Final velocity [itex]\Omega^2 = \omega^2 + 2\alpha \theta[/itex], which gives [itex]\Omega = 8.68[/itex]. But the correct answer is apparently 12.3.