Question

A recording disc rotates steadily at 45 rev/minute on a table. When a small mass of 0.02 kg is dropped gently on the disc at a distance of 0.04 m from its axis and stuck to the disc, the rate of revolution falls to 36 min. The moment of inertia of the disc about its centre is:-

एक रिकार्डिंग चकती 45 चक्कर/मिनट की एकसमान दर से एक मेज पर घूम रही है। जब एक छोटा पिण्ड जो 0.02 किग्रा का है चकती पर गिराया जाता है उसकी अक्ष से 0.04 m की दूरी पर तथा यह पिण्ड चकती से चिपक जाता है तो घूमने की दर 36 चक्कर/मिनट हो जाती है। चकती का अपनी अक्ष के पारित जड़त्व आघूर्ण होगाः-

• A. 1.3x10-4 kgxm2
• B. 1.3x10-5 kgxm2
• C. 1.3x10-3 kgxm2
• D. 1.3x10-2kgxm2

Answer 1

Answer: (A)

1.3x10-4 kgxm2

Answer 2

The problem involves a rotating disc and a mass dropped onto it, leading to a change in the rate of revolution. Since there are no external torques acting on the system (disc + mass), the total angular momentum of the system is conserved.

1. Identify Given Values:

• Initial angular speed of the disc, $\omega_1 = 45 \text{ rev/minute}$
• Final angular speed of the disc + mass, $\omega_2 = 36 \text{ rev/minute}$
• Mass dropped, $m = 0.02 \text{ kg}$
• Distance of the mass from the axis, $r = 0.04 \text{ m}$
• Let $I$ be the moment of inertia of the disc about its center (what we need to find).

2. Calculate Moments of Inertia:

• Initial Moment of Inertia ($I_1$): Before the mass is dropped, only the disc is rotating. So, $I_1 = I$.
• Final Moment of Inertia ($I_2$): After the mass is dropped and sticks to the disc, the system consists of the disc and the point mass. The moment of inertia of a point mass $m$ at a distance $r$ from the axis is $mr^2$. So, $I_2 = I + mr^2$.

3. Apply Conservation of Angular Momentum: According to the principle of conservation of angular momentum, the initial angular momentum ($L_1$) equals the final angular momentum ($L_2$): $L_1 = L_2$ $I_1 \omega_1 = I_2 \omega_2$ Substitute the expressions for $I_1$ and $I_2$: $I \omega_1 = (I + mr^2) \omega_2$

4. Solve for $I$: Expand the equation: $I \omega_1 = I \omega_2 + mr^2 \omega_2$ Rearrange to group terms with $I$: $I \omega_1 - I \omega_2 = mr^2 \omega_2$ $I (\omega_1 - \omega_2) = mr^2 \omega_2$ $I = \frac{mr^2 \omega_2}{\omega_1 - \omega_2}$

5. Substitute the Numerical Values: It's not necessary to convert angular speeds from rev/minute to rad/s, as the units will cancel out in the ratio $\frac{\omega_2}{\omega_1 - \omega_2}$.

• Calculate $mr^2$: $mr^2 = (0.02 \text{ kg}) \times (0.04 \text{ m})^2$ $mr^2 = 0.02 \times 0.0016$ $mr^2 = 0.000032 \text{ kg m}^2 = 3.2 \times 10^{-5} \text{ kg m}^2$

• Substitute all values into the equation for $I$: $I = \frac{(3.2 \times 10^{-5}) \times 36}{45 - 36}$ $I = \frac{(3.2 \times 10^{-5}) \times 36}{9}$ $I = (3.2 \times 10^{-5}) \times 4$ $I = 12.8 \times 10^{-5} \text{ kg m}^2$ $I = 1.28 \times 10^{-4} \text{ kg m}^2$

6. Compare with Options: The calculated value $1.28 \times 10^{-4} \text{ kg m}^2$ is approximately $1.3 \times 10^{-4} \text{ kg m}^2$.

The final answer is $\boxed{\text{1.3x10-4 kgxm2}}$.

Explanation of the solution: The problem is solved using the principle of conservation of angular momentum. The initial angular momentum of the disc ($I\omega_1$) is equated to the final angular momentum of the disc-plus-mass system ($(I+mr^2)\omega_2$). By substituting the given values for mass, radius, and angular speeds, the moment of inertia of the disc ($I$) is calculated.