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 \omega1 and \omega2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:

  • Option 1)

    \frac{1} {2}\text{I}\left( {\omega _1 + \omega _2 } \right)^2

  • Option 2)

    \frac{1} {4}\text{I}\left( {\omega _1 - \omega _2 } \right)^2

  • Option 3)

    \text{I}\left( {\omega _1 - \omega _2 } \right)^2

  • Option 4)

    \frac{\text{I}} {\text{8}}\left( {\omega _1 - \omega _2 } \right)^2

 

Answers (1)

Resolving power \varphi \frac{1}{\lambda }

\therefore \frac{R_{1}}{R_{2}}=\frac{\lambda _{2}}{\lambda_{1}}=3:2

 

Law of conservation of angular moment -

vec{	au }= frac{vec{dL}}

- wherein

If net torque is zero

i.e. frac{vec{dL}}= 0

vec{L}= constant

angular momentum is conserved only when external torque is zero .

 

Iw_{1}+Iw_{2} = 2Iw \Rightarrow w=\frac{w_{1}+w_{2}}{2}

\left ( K.E \right )_{i} = \frac{1}{2}Iw_{1}^{2}+\frac{1}{2}w_{2}^{2}

\left (K.E \right )_{f} = \frac{1}{2}\times 2Iw^{2} = I \left ( \frac{w_{1+w_{2}}}{2} \right )^{2}

Loss in K.E = \frac{1}{4}I\left ( w_{1}-w_{2} \right )^{2}


Option 1)

\frac{1} {2}\text{I}\left( {\omega _1 + \omega _2 } \right)^2

This option is incorrect

Option 2)

\frac{1} {4}\text{I}\left( {\omega _1 - \omega _2 } \right)^2

This option is correct.

Option 3)

\text{I}\left( {\omega _1 - \omega _2 } \right)^2

This option is correct

Option 4)

\frac{\text{I}} {\text{8}}\left( {\omega _1 - \omega _2 } \right)^2

This option is incorrect

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