A constant power is suppied to a rotating disc. The relationship between the angular velocity `(omega)` of the disc and number of rotations `(n)` made by the disc is governed by

Options

A. `(1)/(3)`
B. `(3)/(2)`
C. `(2)/(3)`
D. 2

Answer

To solve the problem of finding the relationship between the angular velocity \((\omega)\) of a rotating disc and the number of rotations \((n)\) made by the disc when a constant power is supplied, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Power and Energy Relationship**:
- Given that a constant power \(P\) is supplied to the rotating disc, the energy \(E\) acquired by the disc over time \(t\) can be expressed as:
\[ E = P \cdot t \quad \text{(Equation 1)} \]

2. **Express the Rotational Energy**:
- The energy of rotation for a disc with moment of inertia \(I\) and angular velocity \(\omega\) is given by:
\[ E = \frac{1}{2} I \omega^2 \quad \text{(Equation 2)} \]

3. **Equate the Two Energy Expressions**:
- Setting the two expressions for energy equal to each other gives:
\[ P \cdot t = \frac{1}{2} I \omega^2 \]

4. **Differentiate with Respect to Time**:
- Differentiate both sides with respect to time \(t\):
\[ \frac{d}{dt}(P \cdot t) = \frac{d}{dt}\left(\frac{1}{2} I \omega^2\right) \]
- This results in:
\[ P = \frac{1}{2} I \cdot 2\omega \frac{d\omega}{dt} \implies P = I \omega \frac{d\omega}{dt} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{d\omega}{dt} = \frac{P}{I \omega} \]

6. **Express in Terms of Angular Position**:
- Multiply and divide by \(d\theta\) (the change in angular position):
\[ \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \frac{P}{I \omega} \]
- Since \(\frac{d\theta}{dt} = \omega\), we have:
\[ \omega \frac{d\omega}{d\theta} = \frac{P}{I} \]

7. **Separate Variables**:
- Rearranging gives:
\[ \omega^2 d\omega = \frac{P}{I} d\theta \]

8. **Integrate Both Sides**:
- Integrate from \(0\) to \(\omega\) on the left and from \(0\) to \(2n\pi\) on the right:
\[ \int_0^\omega \omega^2 d\omega = \frac{P}{I} \int_0^{2n\pi} d\theta \]
- This results in:
\[ \frac{\omega^3}{3} = \frac{P}{I} (2n\pi) \]

9. **Solve for \(\omega\)**:
- Rearranging gives:
\[ \omega^3 = \frac{6n\pi P}{I} \]
- Taking the cube root:
\[ \omega = \left(\frac{6n\pi P}{I}\right)^{1/3} \]

10. **Establish the Relationship**:
- From the above expression, we can see that \(\omega\) is proportional to \(n^{1/3}\):
\[ \omega \propto n^{1/3} \]

### Conclusion:
The relationship between the angular velocity \(\omega\) of the disc and the number of rotations \(n\) made by the disc is given by:
\[ \omega \propto n^{1/3} \]