Question

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ${\omega }_1$ and ${\omega }_2$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss coinciding of energy during this process is:

• A. $\frac{1}{8}({\omega }_1-{\omega }_2{)}^2$
• B. $\frac{1}{2}I({\omega }_1+{\omega }_2{)}^2$
• C. $\frac{1}{4}I({\omega }_1-{\omega }_2{)}^2$
• D. $I({\omega }_1-{\omega }_2{)}^2$

Answer 1

The correct option is C $\frac{1}{4}I({\omega }_1-{\omega }_2{)}^2$

Let the angular velocity of the combination be W
conservation of angular momentum:
$I{w}_1+I{w}_2=(I+I)w$
$\to \frac{1}{2}(w_1+w_2)$
Initial kinetic energy $ki=\frac{1}{2}I{w}_1^2+\frac{1}{2}I{w}_2^2$
Final kinectic energy $k_f=\frac{1}{2}\left(2I\right)w^2$
$\frac{I}{4}(w_1+w_2{)}^2$
$\therefore$ Loss in energy $\delta k=ki-k_f$
$=\frac{1}{2}I(w_1^2+w_2^2)+\frac{1}{4}(w_1^2+w_2^2+2w_1{w}_2)$
$=\frac{I}{4}[w_1^2+w_2^2-2w_1{w}_2]$
$=\frac{I}{4}(w_1-w_2{)}^2$