Question

Two discs of moment of interia $I_1$ and $I_2$ are rotating with angular velocities ${\omega }_1$ and ${\omega }_2$ along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate in combination along the same axis the rotational kinetic energy of system will be :

• A. $\frac{(I_1{\omega }_1+I_2{\omega }_2)}{2(I_1+I_2{)}^2}$
• 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. $\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2)}$

Answer 1

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

There is no external torque applied, hence angular momentum is conserved.
$L_i=L_f$

$\therefore I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$

$\Rightarrow \omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

$K_i=\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2)$

$K_f=\frac{1}{2}(I_1+I_2){\omega }^2=\frac{1}{2}(I_1+I_2)(\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}{)}^2=\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{2(I_1+I_2)}$