While spinning down from 500 rpm to rest, a flywheel does $3.9 \mathrm {~kJ}$ of work. This flywheel is in the shape of a solid uniform disk of radius 1.2 m. What is the mass of this flywheel?

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Quizplus · Accepted Answer

The correct answer is A.The work done by the flywheel is equal to the change in its rotational kinetic energy. The rotational kinetic energy of a solid uniform disk is given by:$$K = \frac{1}{2} I \omega^2$$where:* K is the rotational kinetic energy* I is the moment of inertia* ω is the angular velocityThe moment of inertia of a solid uniform disk is given by:$$I = \frac{1}{2} MR^2$$where:* M is the mass* R is the radiusSubstituting the given values into the equation for rotational kinetic energy, we get:$$K = \frac{1}{2} \left(\frac{1}{2} MR^2ight) \omega^2 = \frac{1}{4} MR^2 \omega^2$$The change in rotational kinetic energy is:$$\Delta K = K_f - K_i = \frac{1}{4} MR^2 \omega_f^2 - \frac{1}{4} MR^2 \omega_i^2$$where:* ω_i is the initial angular velocity* ω_f is the final angular velocitySubstituting the given values into the equation for the change in rotational kinetic energy, we get:$$\Delta K = \frac{1}{4} M(1.2 	ext{ m})^2 (0 	ext{ rad/s})^2 - \frac{1}{4} M(1.2 	ext{ m})^2 (500 	ext{ rpm})^2$$Converting the initial angular velocity from rpm to rad/s:$$\omega_i = 500 	ext{ rpm} \left(\frac{2\pi 	ext{ rad}}{60 	ext{ s}}ight) = 52.36 	ext{ rad/s}$$Substituting the converted value into the equation for the change in rotational kinetic energy, we get:$$\Delta K = \frac{1}{4} M(1.2 	ext{ m})^2 (0 	ext{ rad/s})^2 - \frac{1}{4} M(1.2 	ext{ m})^2 (52.36 	ext{ rad/s})^2$$$$\Delta K = -3.9 	ext{ kJ}$$Solving for the mass, we get:$$M = \frac{4\Delta K}{R^2 \omega_i^2} = \frac{4(-3.9 	ext{ kJ})}{(1.2 	ext{ m})^2 (52.36 	ext{ rad/s})^2} = 4.0 	ext{ kg}$$Therefore, the mass of the flywheel is 4.0 kg.

While Spinning Down from 500 Rpm to Rest, a Flywheel 3.9 kJ3.9 \mathrm {~kJ}3.9 kJ

While Spinning Down from 500 Rpm to Rest, a Flywheel $3.9 \mathrm {~kJ}$