To solve this problem, we need to use the principle of conservation of angular momentum. Since no external torques act on the system (disk + rod), the total angular momentum remains constant.

Step 1: Define Known Quantities

1. Disk:

   • Angular velocity, $\omega_{\text{disk}} = 2.4 \, \text{rev/s}$.
   • Convert to radians per second: $\omega_{\text{disk}} = 2.4 \times 2\pi \, \text{rad/s} = 4.8\pi \, \text{rad/s}$.

2. Rod:

   • Initially, it is not rotating, so its angular velocity is $\omega_{\text{rod}} = 0$.

3. Masses and Dimensions:

   • The rod and disk have the same mass, $m$, and the rod length equals the disk diameter.

Step 2: Calculate Moments of Inertia

1. Moment of Inertia of the Disk (rotating about its center): $I_{\text{disk}} = \frac{1}{2} m R^2$ where $R$ is the radius of the disk.

2. Moment of Inertia of the Rod (rotating about its center): $I_{\text{rod}} = \frac{1}{12} m L^2$ where $L$ is the length of the rod. Since $L$ equals the diameter of the disk, $L = 2R$. Substitute $L = 2R$ into the equation: $I_{\text{rod}} = \frac{1}{12} m (2R)^2 = \frac{1}{12} m \cdot 4R^2 = \frac{1}{3} m R^2$

3. Total Moment of Inertia of the System: When the rod is dropped onto the disk, they rotate together as a single object, so we add their moments of inertia: $I_{\text{total}} = I_{\text{disk}} + I_{\text{rod}} = \frac{1}{2} m R^2 + \frac{1}{3} m R^2 = \frac{5}{6} m R^2$

Step 3: Apply Conservation of Angular Momentum

Let $\omega_{\text{final}}$ be the angular velocity of the combined system (disk + rod) after the rod is dropped. Since angular momentum is conserved: $I_{\text{disk}} \cdot \omega_{\text{disk}} = I_{\text{total}} \cdot \omega_{\text{final}}$ Substitute $I_{\text{disk}} = \frac{1}{2} m R^2$, $I_{\text{total}} = \frac{5}{6} m R^2$, and $\omega_{\text{disk}} = 4.8\pi$: $\frac{1}{2} m R^2 \cdot 4.8\pi = \frac{5}{6} m R^2 \cdot \omega_{\text{final}}$

Now, cancel $m R^2$ from both sides: $\frac{1}{2} \cdot 4.8\pi = \frac{5}{6} \cdot \omega_{\text{final}}$

Multiply both sides by $\frac{6}{5}$: $\omega_{\text{final}} = \frac{6}{5} \cdot \frac{1}{2} \cdot 4.8\pi = \frac{6 \cdot 4.8\pi}{10} = 2.88\pi \, \text{rad/s}$

Answer

The angular velocity of the combined system is: $\omega_{\text{final}} = 2.88\pi \, \text{rad/s} \approx 9.04 \, \text{rad/s}$

Would you like more details or have any questions?

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Related Questions:

1. How would the result change if the rod had twice the mass of the disk?
2. What happens to the angular velocity if the disk were initially at rest and only the rod rotated?
3. How would the outcome differ if the rod were dropped off-center on the disk?
4. What if the rod were shorter than the diameter of the disk? How would the inertia calculation change?
5. How would the problem change if friction was not negligible between the rod and the disk?

Tip: Always check units, especially when converting from revolutions per second to radians per second.