Question

Two discs of same moment of inertia rotating about their regular axis passing through center 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. $\dfrac{1}{8}{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}$
B. $\dfrac{1}{2}I{{({{\omega }_{1}}+{{\omega }_{2}})}^{2}}$
C. $\dfrac{1}{4}I{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}$
D. $I{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}$

Answer 1

First understand that moment of inertia depends on the axis of rotation, shape and size of the body. Apply the law conservation angular momentum to get a relationship between the angular velocity of the combined system and the individual angular velocities. Using the formula of kinetic energy find out the expression for initial and final kinetic energy and hence the loss.

Complete step-by-step solution -
Let I be the moment of inertia of two discs about the regular axis passing through the center and perpendicular to the plane of disc.${\omega _1}$and${\omega _2}$be their angular velocities.
Therefore, the angular momentum of the first disc is ${\rm I}{\omega _1}$and the angular momentum of the second disc is ${\rm I}{\omega _2}$.
Initial angular momentum = ${\rm I}{\omega _1} + {\rm I}{\omega _2}$
Let $\omega$ be the angular velocity of the combined system.
Therefore, final angular momentum = $({\rm I} + {\rm I})\omega = 2{\rm I}\omega$
According to the law of conservation of angular momentum,
Initial angular momentum = final angular momentum
\eqalign{ & \therefore {\rm I}{\omega _1} + {\rm I}{\omega _2} = 2{\rm I}\omega \cr & \therefore \omega = \dfrac{{{\omega _1} + {\omega _2}}}{2} \cr}
Initial kinetic energy = $\dfrac{1}{2}{\rm I}{\omega _1}^2 + \dfrac{1}{2}{\rm I}{\omega _2}^2 = \dfrac{I}{2}({\omega _1}^2 + {\omega _2}^2)$
Final kinetic energy = $\dfrac{1}{2}(2{\rm I}){\omega ^2} = {\rm I}{\omega ^2}$
Loss in kinetic energy $({E_L})$ = Final kinetic energy – Initial kinetic energy
\eqalign{ & {E_L} = \dfrac{I}{2}({\omega _1}^2 + {\omega _2}^2) - {\rm I}{\omega ^2} \cr & {E_L} = \dfrac{I}{2}({\omega _1}^2 + {\omega _2}^2) - {\rm I}{(\dfrac{{{\omega _1} + {\omega _2}}}{2})^2} \cr & {E_L} = \dfrac{I}{4}(2{\omega _1}^2 + 2{\omega _2}^2 - {\omega _1}^2 - 2{\omega _1}{\omega _2} - {\omega _2}^2) \cr & {E_L} = \dfrac{I}{4}({\omega _1}^2 + {\omega _2}^2 - 2{\omega _1}{\omega _2}) \cr & {E_L} = \dfrac{I}{4}{({\omega _1} - {\omega _2})^2} \cr}
The expression for loss coinciding with energy during this process is $\dfrac{I}{4}{({\omega _1} - {\omega _2})^2}$. Therefore options (C) is the correct option.

Note: Note that angular momentum is conserved when external torque acting on a system is zero. The moment of inertia is different along the different axis of rotation. The moment of inertia also depends on the shape and size of the object.