JEE Physics 2026 Question with Solution

Given: $M = 3\,kg$, $R = 5\,m$, $m = 3\,kg$, $h = 3\,m$, $g = 10\,m\,s^{-2}$

Find: Kinetic energy of the flywheel after the mass descends by $3\,m$.

Concept: The loss in gravitational potential energy of the falling mass is converted into translational kinetic energy of the mass and rotational kinetic energy of the flywheel.

Moment of inertia of the flywheel is

$$I=\frac{1}{2}MR^2$$

Apply energy conservation:

$$mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$$

Loss in potential energy:

$$mgh = 3 \times 10 \times 3 = 90\ \text{J}$$

Relate linear and angular speed:

$$v=R\omega \Rightarrow \omega=\frac{v}{R}$$

Substitute moment of inertia:

$$I=\frac{1}{2}(3)(5)^2=\frac{75}{2}$$

Now put these into the energy equation:

$$90=\frac{1}{2}(3)v^2+\frac{1}{2}\left(\frac{75}{2}\right)\left(\frac{v}{5}\right)^2$$ $$90=\frac{3}{2}v^2+\frac{75}{4}\cdot\frac{v^2}{25}$$ $$90=\frac{3}{2}v^2+\frac{3}{4}v^2$$ $$90=\frac{9}{4}v^2$$ $$v^2=40$$

Kinetic energy of the flywheel:

$$K_{\text{flywheel}}=\frac{1}{2}I\omega^2$$ $$K_{\text{flywheel}}=\frac{1}{2}\cdot\frac{75}{2}\cdot\frac{40}{25}$$ $$K_{\text{flywheel}}=30\ \text{J}$$

Therefore, the kinetic energy of the flywheel is $30\,\text{J}$.

Given: A falling mass of $3\,kg$ is attached to a flywheel of mass $3\,kg$ and radius $5\,m$.

Find: The rotational kinetic energy gained by the flywheel.

The total gravitational potential energy lost is

$$90\ \text{J}$$

This energy is shared between:

1. Translational kinetic energy of the hanging mass
2. Rotational kinetic energy of the flywheel

Because the string does not slip,

$$v=R\omega$$

so both kinetic energy terms can be written using the same speed variable.

Using

$$I=\frac{1}{2}MR^2=\frac{75}{2}$$

we get

$$\frac{1}{2}I\omega^2=\frac{1}{2}\cdot\frac{75}{2}\cdot\frac{v^2}{25}=\frac{3}{4}v^2$$

Also,

$$\frac{1}{2}mv^2=\frac{3}{2}v^2$$

Hence the energy equation becomes

$$90=\frac{3}{2}v^2+\frac{3}{4}v^2=\frac{9}{4}v^2$$

which gives

$$v^2=40$$

Then the flywheel kinetic energy is

$$\frac{3}{4}v^2=\frac{3}{4}\times 40=30\ \text{J}$$

Therefore, the required kinetic energy is $30\,\text{J}$.