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?
A) 4.0 kg
B) 3.4 kg
C) 4.6 kg
D) 5.2 kg

Question

Quizplus · Accepted Answer

The work done by the flywheel is equal to its change in kinetic energy. The kinetic energy of a rotating disk is given by:$$K = \frac{1}{2}I\omega^2$$where I is the moment of inertia and $\omega$ is the angular velocity. The moment of inertia of a solid uniform disk is given by:$$I = \frac{1}{2}MR^2$$where M is the mass and R is the radius. Substituting these expressions into the equation for kinetic energy, we get:$$K = \frac{1}{4}MR^2\omega^2$$The change in kinetic energy is:$$\Delta K = K_f - K_i = \frac{1}{4}MR^2(\omega_f^2 - \omega_i^2)$$where $\omega_i$ is the initial angular velocity and $\omega_f$ is the final angular velocity. Substituting the given values, we get:$$\Delta K = \frac{1}{4}M(1.2 \mathrm{~m})^2[(0 \mathrm{~rad/s})^2 - (500 \mathrm{~rpm})(2\pi \mathrm{~rad/min}/60 \mathrm{~s/min})^2]$$$$\Delta K = \frac{1}{4}M(1.44 \mathrm{~m}^2)(1.05 \times 10^4 \mathrm{~rad^2/s^2})$$$$\Delta K = 4.0M \mathrm{~J}$$Setting this equal to the work done by the flywheel, we get:$$4.0M \mathrm{~J} = 3.9 \mathrm{~kJ}$$$$M = \frac{3.9 \mathrm{~kJ}}{4.0 \mathrm{~J/kg}} = 0.975 \mathrm{~kg}$$Therefore, the mass of the flywheel is 0.975 kg, which is closest to choice A) 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 $3.9 \mathrm {~kJ}$