The problem asks for the kinetic energy of a flywheel after a connected mass descends.
The total potential energy lost by the descending mass is converted into the kinetic energy of the mass and the kinetic energy of the rotating flywheel. We assume the flywheel is a solid disk for calculating its moment of inertia.
The energy conservation equation is:
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $Where:
Relationship between linear and angular velocity: $v = R\omega$ or $\omega = v/R$.
Assuming the flywheel is a solid disk, its moment of inertia ($I$) is:
$ I = \frac{1}{2}MR^2 $ $ I = \frac{1}{2}(3\text{ kg})(5\text{ m})^2 = \frac{1}{2}(3)(25) = 37.5\text{ kg m}^2 $Substitute $\omega = v/R$ and $I = 37.5\text{ kg m}^2$ into the energy equation:
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 $ $ mgh = \frac{1}{2}mv^2 + \frac{1}{4}MR^2\frac{v^2}{R^2} $ $ mgh = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2 $ $ mgh = \frac{1}{2}v^2\left(m + \frac{1}{2}M\right) $Plug in the values:
$ (3\text{ kg})(10\text{ m/s}^2)(3\text{ m}) = \frac{1}{2}v^2\left(3\text{ kg} + \frac{1}{2}(3\text{ kg})\right) $ $ 90\text{ J} = \frac{1}{2}v^2\left(3 + 1.5\right)\text{ kg} $ $ 90\text{ J} = \frac{1}{2}v^2(4.5\text{ kg}) $ $ 180\text{ J} = v^2(4.5\text{ kg}) $ $ v^2 = \frac{180}{4.5}\text{ m}^2/\text{s}^2 = 40\text{ m}^2/\text{s}^2 $The kinetic energy of the flywheel ($KE_{wheel}$) is:
$ KE_{wheel} = \frac{1}{2}I\omega^2 = \frac{1}{2}I\left(\frac{v}{R}\right)^2 = \frac{1}{2}\frac{I}{R^2}v^2 $We know $I = 37.5\text{ kg m}^2$ and $R = 5\text{ m}$, so $\frac{I}{R^2} = \frac{37.5}{25} = 1.5\text{ kg}$
$ KE_{wheel} = \frac{1}{2}(1.5\text{ kg})(40\text{ m}^2/\text{s}^2) $ $ KE_{wheel} = \frac{1}{2}(60\text{ J}) $ $ KE_{wheel} = 30\text{ J} $A thin uniform rod ($X$) of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is __________.
($g = \text{gravitational acceleration}$)

The pulley shown in figure is made using a thin rim and two rods of length equal to diameter of the rim. The rim and each rod have a mass of $M$. Two blocks of mass of $M$ and $m$ are attached to two ends of a light string passing over the pulley, which is hinged to rotate freely in vertical plane about its center. The magnitudes of the acceleration experienced by the blocks is __________
(assume no slipping of string on pulley).
A thin uniform rod ($X$) of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is __________.
($g = \text{gravitational acceleration}$)
