Two discs of moment of inertia $I_1=4 \mathrm{~kg} \mathrm{~m}^2$ and $I_2=2 \mathrm{~kg} \mathrm{~m}^2$, about their central axes & normal to their planes, rotating with angular speeds $10 \mathrm{~rad} / \mathrm{s}$ & $4 \mathrm{~rad} / \mathrm{s}$ respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is _________ J.

<p>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:</p><p>First, because angular momentum before and after they touch must be the same, we have:</p><p>$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$</p><p>where:</p><ul><li>$I_1$ and $I_2$ are the moments of inertia for the two discs.</li><li>$\omega_1$ and $\omega_2$ are their angular speeds before contact.</li><li>$\omega_0$ is their common angular speed after contact.</li></ul><p>Plugging in the given values, we find that:</p><p>$\omega_0 = 8 \mathrm{rad/s}$</p><p>Next, to calculate the loss in kinetic energy, we first find the total kinetic energy before and after they touch:</p><p>Before: $E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = 216 \,\mathrm{J}$</p><p>After: $E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = 192 \,\mathrm{J}$</p><p>The loss in kinetic energy ($\Delta E$) is the difference:</p><p>$\Delta E = E_1 - E_2 = 24 \,\mathrm{J}$</p><p>So, when the two discs are brought together, the system loses 24 J of kinetic energy.</p>