Question 488: Centre of Mass and Rotational Motion

Question

Two discs of moment of inertia ' $\mathrm{I}_1$ ' and ' $\mathrm{I}_2$ ' and angular speeds ' $\omega_1$ ' and ' $\omega_2$ ' are rotating along the collinear axes passing through their centre of mass and perpendicular to their plane. If the two discs are made to rotate together along the same axis. The rotational kinetic energy of the system will be

Accepted Answer

Step 1: Conservation principle
When the two discs are suddenly coupled (say by clamping them together along the same axis),
Angular momentum is conserved
(because external torque about the axis is zero).
Kinetic energy is not necessarily conserved
(some energy may be lost as heat/friction during coupling).
Step 2: Total angular momentum before coupling
For disc 1:
$ L_1 = I_1 \omega_1 $
For disc 2:
$ L_2 = I_2 \omega_2 $
So, total angular momentum:
$ L = I_1 \omega_1 + I_2 \omega_2 $
Step 3: Common angular velocity after coupling
When they rotate together: Let the final angular velocity = $ \omega $.
Moment of inertia = $ I_1 + I_2 $.
From conservation of angular momentum:
$ (I_1 + I_2)\, \omega = I_1 \omega_1 + I_2 \omega_2 $
$ \omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} $
Step 4: Final rotational kinetic energy
$ K = 	frac{1}{2} (I_1 + I_2) \, \omega^2 $
Substitute $\omega$ from above:
$ K = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} ight)^2 $
$ K = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2 (I_1 + I_2)} $
Final Answer:
$ \boxed{\frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2 (I_1 + I_2)}} $
That corresponds to
Option D
.

Answer

$\frac{I_1 \omega_1+I_2 \omega_2}{2\left(I_1+I_2\right)^2}$

Answer

$\frac{\left(I_1 \omega_1-I_2 \omega_2\right)^2}{2\left(I_1+I_2\right)}$

Answer

$\quad \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{2\left(I_1-I_2\right)}$

Answer

$\frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{2\left(I_1+I_2\right)}$

Centre of Mass and Rotational Motion

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