Question

Block of mass \(3\,\text{kg}\) is connected to a flywheel of mass \(3\,\text{kg}\) and radius \(5\,\text{m}\) through a massless string wrapped around the flywheel. Find kinetic energy (in J) of flywheel when block descends by \(3\,\text{m}\).

Hint:

In block–pulley–flywheel systems, always split energy into {translational} and {rotational} parts and use the no-slip condition \(v = \omega R\).

Answer 1

Correct Answer: 30

Concept:
When a block descends while rotating a flywheel:

Loss of gravitational potential energy converts into kinetic energy.
Kinetic energy is shared between:

Translational kinetic energy of the block
Rotational kinetic energy of the flywheel

For a flywheel (solid disc), moment of inertia: \[ I = \frac{1}{2}MR^2 \]

Step 1: Loss of Gravitational Potential Energy
\[ \Delta U = mgh = 3 \times 10 \times 3 = 90\,\text{J} \]
Step 2: Express Kinetic Energies
Let the speed of the block be \(v\). Since the string does not slip: \[ v = \omega R \] Kinetic energy of block:
\[ K_{\text{block}} = \frac{1}{2}mv^2 \] Kinetic energy of flywheel:
\[ K_{\text{flywheel}} = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{4}Mv^2 \]
Step 3: Apply Energy Conservation
\[ 90 = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2 \] Substitute \(m = 3\,\text{kg}\), \(M = 3\,\text{kg}\): \[ 90 = \left(\frac{1}{2}\times 3 + \frac{1}{4}\times 3\right)v^2 \] \[ 90 = \left(1.5 + 0.75\right)v^2 = 2.25v^2 \] \[ v^2 = 40 \]
Step 4: Flywheel Kinetic Energy
\[ K_{\text{flywheel}} = \frac{1}{4}Mv^2 = \frac{1}{4}\times 3 \times 40 = 30\,\text{J} \] \[ \boxed{K_{\text{flywheel}} = 30\,\text{J}} \]