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 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:$$K = \frac{1}{2} \left(\frac{1}{2} MR^2\right) \omega^2$$$$K = \frac{1}{4} MR^2 \omega^2$$The work done by the flywheel is equal to the change in its rotational kinetic energy, so:$$W = K_f - K_i$$$$W = \frac{1}{4} MR^2 \omega_f^2 - \frac{1}{4} MR^2 \omega_i^2$$$$W = \frac{1}{4} MR^2 (\omega_f^2 - \omega_i^2)$$Substituting the given values into the equation for work, we get:$$3.9 \mathrm {~kJ} = \frac{1}{4} M(1.2 \mathrm {~m})^2 (0 - (500 \mathrm {~rpm})^2)$$$$3.9 \mathrm {~kJ} = \frac{1}{4} M(1.44 \mathrm {~m}^2) (-(500 \mathrm {~rpm})^2)$$$$3.9 \mathrm {~kJ} = \frac{1}{4} M(1.44 \mathrm {~m}^2) (-250000 \mathrm {~rpm}^2)$$$$3.9 \mathrm {~kJ} = -937500 \mathrm {~kg~m}^2/\mathrm {~s}^2$$Solving for M, we get:$$M = \frac{3.9 \mathrm {~kJ}}{-937500 \mathrm {~kg~m}^2/\mathrm {~s}^2}$$$$M = 4.16 \mathrm {~kg}$$Therefore, the mass of the flywheel is 4.16 kg, which is closest to choice A, 4.0 kg.

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

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