Question

A flywheel having mass \(3\,\text{kg}\) and radius \(5\,\text{m}\) 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 a \(3\,\text{kg}\) mass. The mass is kept initially and released. Kinetic energy of the flywheel when the mass descends by \(3\,\text{m}\) is _______ J. \((g=10\,\text{m s}^{-2})\) Given: \[ M = 3\,\text{kg}, \quad R = 5\,\text{m}, \quad m = 3\,\text{kg}, \quad h = 3\,\text{m}, \quad g = 10\,\text{m s}^{-2} \]

Hint:

Always apply energy conservation in string–pulley–flywheel systems and relate linear and angular speeds using \(v=R\omega\).

Answer 1

Correct Answer: 30

Concept: The loss in gravitational potential energy of the falling mass is converted into:
Translational kinetic energy of the mass
Rotational kinetic energy of the flywheel
Moment of inertia of a flywheel (solid disc) is: \[ I=\frac{1}{2}MR^2 \]
Step 1: Write energy conservation equation Loss in potential energy: \[ mgh = 3 \times 10 \times 3 = 90\ \text{J} \] This equals total kinetic energy: \[ mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \]
Step 2: Relate linear and angular velocity \[ v=R\omega \Rightarrow \omega=\frac{v}{R} \]
Step 3: Substitute moment of inertia \[ I=\frac{1}{2}(3)(5)^2=\frac{75}{2} \]
Step 4: Substitute into energy equation \[ 90=\frac{1}{2}(3)v^2+\frac{1}{2}\left(\frac{75}{2}\right)\left(\frac{v}{5}\right)^2 \] \[ 90=\frac{3}{2}v^2+\frac{75}{4}\cdot\frac{v^2}{25} \] \[ 90=\frac{3}{2}v^2+\frac{3}{4}v^2 =\frac{9}{4}v^2 \] \[ v^2=40 \]
Step 5: Calculate kinetic energy of the flywheel \[ K_{\text{flywheel}}=\frac{1}{2}I\omega^2 =\frac{1}{2}\cdot\frac{75}{2}\cdot\frac{40}{25} \] \[ K_{\text{flywheel}}=30\ \text{J} \] Final Answer: \[ \boxed{30\ \text{J}} \]