Question

Two discs of moment of inertia $I_1$ and $I_2$ having angular speed ${\omega }_1$ and ${\omega }_2$ are rotating along collinear axis 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 $\text{KE}$ of system will be

• A. $\frac{I_1{\omega }_1+I_2{\omega }_2}{2(I_1+I_2)}$
• B. $\frac{(I_1+I_2)({\omega }_1+{\omega }_2{)}^2}{2}$
• C. $\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{2(I_1+I_2)}$
• D. $\text{None of these}$

Answer 1

The correct option is C $\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{2(I_1+I_2)}$

By the law of conservation of angular momentum, $I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$
Angular velocity of system , $\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$
Rotational kinetic energy
$=\frac{1}{2}(I_1+I_2){\omega }^2=\frac{1}{2}(I_1+I_2){\left(\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}\right)}^2$
$=\frac{(I_1\omega +I_2{\omega }_2{)}^2}{2(I_1+I_2)}$