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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)

best_answer

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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