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$\cdot$ The flywheel of a motor has a mass of 300.0 $\mathrm{kg}$ and a moment of inertia of 580 $\mathrm{kg} \cdot \mathrm{m}^{2} .$ The motor develops a constant torque of $2000.0 \mathrm{N} \cdot \mathrm{m},$ and the flywheel starts from rest. (a) What is the angular acceleration of the flywheel? (b) What is its angular velocity after it makes 4.00 revolutions? (c) How much work is done by the motor during the first 4.00 revolutions?

a) 3.45 $\mathrm{rad} / \mathrm{s}^{2}$b) 13.2 $\mathrm{rad} / \mathrm{s}$c) $5.0529 \times 10^{4} \mathrm{J}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

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University of Washington

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So here the work is going to be equal to the torque times, the change and angular displacement. We know that the mass is going to be equal to 300 kilograms. We know that the moment of inertia equals 580 kilogram meters squared, and we know that the torque is equaling 2,000 Newton meters. So we can say that for party. In order to find the angular acceleration, we find that the sum of the torque is equaling the moment of inertia, times the angular acceleration. So the angular acceleration will simply be the network divided by the moment of inertia. So 2,000 divided by 580. And this is giving us 3.45 radiance per second squared now to find bee the change and angular displacement. Ah, rather, to find me that the final angular velocity this will be the we have to use angular kinnah matics. So we know that the angular displacement is equaling toe four revolutions. Or we can say eight pi radiance because there are two pyre. Etienne's for everyone revolution. And to find the final angular velocity, we can say that a mega final squared equals Omega Initial squared, plus two times Alfa Times, Delta Theta. We know that Omega initial the initial angular velocity squared is going to be zero. So we can say that Omega Final equals the square root of two times alpha times Delta theta. And this will equal the square root of two times 3.45 Time's a pie, and this is giving us 13 point to radiance per second. And for part, see, it's asking us for the work done. So the work will be simply the again the torque times, the change and angular displacement. So this would be 2,000 times 13. Rather, my apologies a pie. And this is giving us 50,000 265 jewels of work. So this would be your answer part. See your answer for part B and your answer for party. That is the end of the solution. Thank you for watching

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