QJEEM24Phcf75: JEE Main 2024 Physics - Class 11 JEEM24Phcf75 Solution: Two discs of moment of inertia and , about their central axes & …

To find the loss in kinetic energy when two spinning discs are brought together, we use the principle of conservation of angular momentum and the formula for kinetic energy. Here's how:

First, because angular momentum before and after they touch must be the same, we have:

$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$

where:

• $I_1$ and $I_2$ are the moments of inertia for the two discs.
• $\omega_1$ and $\omega_2$ are their angular speeds before contact.
• $\omega_0$ is their common angular speed after contact.

Plugging in the given values, we find that:

$\omega_0 = 8 \mathrm{rad/s}$

Next, to calculate the loss in kinetic energy, we first find the total kinetic energy before and after they touch:

Before: $E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = 216 \,\mathrm{J}$

After: $E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = 192 \,\mathrm{J}$

The loss in kinetic energy ( $\Delta E$ ) is the difference:

$\Delta E = E_1 - E_2 = 24 \,\mathrm{J}$

So, when the two discs are brought together, the system loses 24 J of kinetic energy.