Two coaxial discs, having moments of inertia I_{1} and \frac{I_{2}}{2}, are rotating with 

respective angular velocities \omega_{1} and  \frac{\omega_{2}}{2}, about their common axis.

They are brought in contact with each other and thereafter they rotate

with a common angular velocity. If E_{f} and E_{i} are the final and initial total 

energies, then (E_{f}-E_{i}) is :

  • Option 1)

    -\frac{I_{1}w_1^{2}}{12}

  • Option 2)

    \frac{I_{1}w_1^{2}}{6}

  • Option 3)

    \frac{3I_{1}w_1^{2}}{8}

  • Option 4)

    -\frac{I_{1}w_1^{2}}{24}

 

Answers (1)

 

Angular momentum -

\vec{L}=\vec{r}\times \vec{p}

- wherein

\vec{L}  represent angular momentum of a moving particle about a point.

it can be calculated  as L=r_1\, P=r\, P_1

r_1 = Length of perpendicular on line of motion

P_1 = component of momentum along perpendicualar to r

 

 

Kinetic energy of a body in combined rotation and translation -

K.E=\frac{1}{2}\, I_{cm}w^{2}+\frac{1}{2}mv_{0}\, ^{2}

- wherein

\frac{1}{2}mv_{0}\, ^{2} =  Translational kinetic energy of centre of mass

 

\frac{1}{2}\, I_{cm}w^{2}  =  Rotational kinetic energy about centre of mass

 

 

F_{ext}=0

So,

L_{i}=L_{f}

L_{i}=I_1\omega _1+\frac{I_1}{2}\frac{\omega _1}{2}

L_{f}=(I_1+\frac{I_1}{2})\omega _f

L_{i}=L_{f}

\frac{5I_1\omega _1}{4}=\frac{3I_1\omega _f}{2}

\omega _f=(\frac{5}{6})\omega_1

K.E_I=\frac{1}{2}I_1\omega _1^{2}+\frac{1}{2}\times \frac{I_1}{2}\times \frac{\omega _1^{2}}{4}

K.E_f=\frac{1}{2}(I_1+\frac{I_1}{2})\omega _f^{2}

\Delta K.E=K.E_f-K.E_I

               =\frac{1}{2}(I_1+\frac{I_1}{2})\omega _f^{2}-(\frac{1}{2}I_1\omega _1^{2}+\frac{1}{2}\times \frac{I_1}{2}\times \frac{\omega _1^{2}}{4})

              =\frac{1}{2}I_1\omega _1^{2}(\frac{25}{24}-\frac{9}{8})

             =-\frac{I_{1}w_1^{2}}{24}

 

 


Option 1)

-\frac{I_{1}w_1^{2}}{12}

Option 2)

\frac{I_{1}w_1^{2}}{6}

Option 3)

\frac{3I_{1}w_1^{2}}{8}

Option 4)

-\frac{I_{1}w_1^{2}}{24}

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