A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string havi

Question

A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is $\_\_\_\_$ $\mathrm{J} .\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Professor PC Thomas Classes · Answer

As the 3 kg mass ($m$) descends by a height $h=3 	ext{ m}$, it loses potential energy $\Delta U = mgh = 3 	imes 10 	imes 3 = 90 	ext{ J}$. This energy is converted into the translational kinetic energy of the mass and the rotational kinetic energy of the flywheel ($M=3 	ext{ kg}$). $\Delta U = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. The moment of inertia of the flywheel (modeled as a disc) is $I = \frac{1}{2}MR^2$. The linear velocity of the mass relates to the angular velocity of the wheel by $v = R\omega$. Substituting these: $90 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2 = \frac{1}{2}(3)v^2 + \frac{1}{4}(3)v^2 = \frac{6}{4}v^2 + \frac{3}{4}v^2 = \frac{9}{4}v^2$. Solving for $v^2$: $v^2 = \frac{90 	imes 4}{9} = 40$. The kinetic energy of the wheel is $\frac{1}{4}Mv^2 = \frac{1}{4} 	imes 3 	imes 40 = 30 	ext{ J}$.