Question 134: Centre of Mass and Rotational Motion

Question

A thin circular disc of mass $$\mathrm{M}$$ and radius $$\mathrm{R}$$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity $$\omega$$. If another disc of same dimensions but of mass $$\mathrm{M} / 2$$ is placed gently on the first disc co-axially, then the new angular velocity of the system is :

Accepted Answer

To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $$L_{	ext{initial}}$$ and $$L_{	ext{final}}$$.
The initial angular momentum of the system is given by:
$$ L_{	ext{initial}} = I_{	ext{initial}} \cdot \omega $$
Here, $$I_{	ext{initial}}$$ is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:
$$ I_{	ext{initial}} = \frac{1}{2} M R^2 $$
Thus,
$$ L_{	ext{initial}} = \left( \frac{1}{2} M R^2 ight) \omega $$
When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity $$\omega'$$. The moment of inertia of the second disc is:
$$ I_{	ext{second}} = \frac{1}{2} \left( \frac{M}{2} ight) R^2 = \frac{1}{4} M R^2 $$
The combined moment of inertia of the system after placing the second disc is:
$$ I_{	ext{final}} = I_{	ext{initial}} + I_{	ext{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2 $$
Thus, the final angular momentum of the system is:
$$ L_{	ext{final}} = I_{	ext{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 ight) \omega' $$
By the conservation of angular momentum:
$$ L_{	ext{initial}} = L_{	ext{final}} $$
This simplifies to:
$$ \left( \frac{1}{2} M R^2 ight) \omega = \left( \frac{3}{4} M R^2 ight) \omega' $$
Solving for $$\omega'$$:
$$ \omega' = \frac{\left( \frac{1}{2} M R^2 ight) \omega}{\left( \frac{3}{4} M R^2 ight)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega $$
So, the new angular velocity of the system is:
$$\omega' = \frac{2}{3} \omega$$
Thus, the correct answer is:
Option D: $$\frac{2}{3} \omega$$

Answer

$$\frac{4}{5} \omega$$

Answer

$$\frac{5}{4} \omega$$

Answer

$$\frac{3}{2} \omega$$

Answer

$$\frac{2}{3} \omega$$

Centre of Mass and Rotational Motion

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