Two discs of moment of inertia I1 and I2 and angular speeds ω1 and ω2 are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational KE of system will be

The correct option is (I₁ω₁ + I₂ω₂)²/2(I₁ + I₂).

Let's derive the expression step-by-step using the principles of rotational dynamics.

Step 1: Understand the initial state of the system
We have two rotating discs:
- The first disc has a moment of inertia $I_1$ and angular speed ${\omega }_1$.
- The second disc has a moment of inertia $I_2$ and angular speed ${\omega }_2$.
Since their axes of rotation are collinear and perpendicular to their planes, their individual angular momenta point along the same line.

The initial angular momentum of the system is the sum of the angular momenta of the two individual discs:
$L_i=I_1{\omega }_1+I_2{\omega }_2$

Step 2: Conservation of Angular Momentum
When the two discs are coupled together to rotate as a single system along the same collinear axis, no external torque acts on the system. Therefore, the total angular momentum of the system is conserved.

Let the common angular speed of the combined system be $\omega$. The total moment of inertia of the combined system is:
$I=I_1+I_2$
The final angular momentum of the combined system is:
$L_f=(I_1+I_2)\omega$

By applying conservation of angular momentum ($L_i=L_f$):
$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$
Solving for the common angular speed $\omega$:
$\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

Step 3: Calculate the Final Rotational Kinetic Energy
The rotational kinetic energy of the combined system is given by:
$E_k=\frac{1}{2}I{\omega }^2$
Substitute the values of $I$ and $\omega$ into the energy equation:
$E_k=\frac{1}{2}(I_1+I_2){\left(\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}\right)}^2$
Simplify the expression by canceling $(I_1+I_2)$ in the numerator and denominator:
$E_k=\frac{1}{2}(I_1+I_2)\frac{{(I_1{\omega }_1+I_2{\omega }_2)}^2}{{(I_1+I_2)}^2}$
$E_k=\frac{{(I_1{\omega }_1+I_2{\omega }_2)}^2}{2(I_1+I_2)}$

Thus, the final rotational kinetic energy of the system is $\frac{{(I_1{\omega }_1+I_2{\omega }_2)}^2}{2(I_1+I_2)}$.