A thin circular metal disc of radius 500.0 mm is set rotating about a central axis normal to its plane. Upon raising its temperature gradually, the radius increases to 507.5 mm. The percentage change in the rotational kinetic energy will be

Options

A. `1.5%`
B. `-1.5%`
C. `3%`
D. `-3%`

Answer

To find the percentage change in the rotational kinetic energy of the disc as its radius increases, we can follow these steps:

### Step 1: Understand the relationship between rotational kinetic energy and moment of inertia
The rotational kinetic energy (K) of a rotating object is given by the formula:
\[ K = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 2: Determine the moment of inertia for a thin circular disc
The moment of inertia \( I \) for a thin circular disc about an axis through its center is given by:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is its radius.

### Step 3: Calculate the initial and final moments of inertia
- Initial radius \( r_1 = 500 \, \text{mm} = 0.5 \, \text{m} \)
- Final radius \( r_2 = 507.5 \, \text{mm} = 0.5075 \, \text{m} \)

Now, calculate the initial and final moments of inertia:
- Initial moment of inertia \( I_1 = \frac{1}{2} m (0.5)^2 = \frac{1}{2} m (0.25) = 0.125 m \)
- Final moment of inertia \( I_2 = \frac{1}{2} m (0.5075)^2 = \frac{1}{2} m (0.2575) = 0.12875 m \)

### Step 4: Use the conservation of angular momentum
Since angular momentum \( L \) is conserved, we have:
\[ L = I \omega \]
Thus, we can express the angular velocity in terms of the moment of inertia:
\[ \omega_1 = \frac{L}{I_1} \quad \text{and} \quad \omega_2 = \frac{L}{I_2} \]

### Step 5: Relate the initial and final angular velocities
From the conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
This implies:
\[ \omega_2 = \frac{I_1}{I_2} \omega_1 \]

### Step 6: Calculate the change in rotational kinetic energy
Now, substituting \( \omega_2 \) into the kinetic energy formula:
- Initial kinetic energy \( K_1 = \frac{1}{2} I_1 \omega_1^2 \)
- Final kinetic energy \( K_2 = \frac{1}{2} I_2 \left( \frac{I_1}{I_2} \omega_1 \right)^2 = \frac{1}{2} I_2 \frac{I_1^2}{I_2^2} \omega_1^2 = \frac{1}{2} \frac{I_1^2}{I_2} \omega_1^2 \)

### Step 7: Find the percentage change in kinetic energy
The change in kinetic energy \( \Delta K = K_2 - K_1 \):
\[ \Delta K = \frac{1}{2} \frac{I_1^2}{I_2} \omega_1^2 - \frac{1}{2} I_1 \omega_1^2 \]
Factoring out \( \frac{1}{2} \omega_1^2 \):
\[ \Delta K = \frac{1}{2} \omega_1^2 \left( \frac{I_1^2}{I_2} - I_1 \right) \]

Now, substituting \( I_1 \) and \( I_2 \):
\[ \Delta K = \frac{1}{2} \omega_1^2 \left( \frac{(0.125 m)^2}{0.12875 m} - 0.125 m \right) \]

Calculating the percentage change:
\[ \frac{\Delta K}{K_1} = \frac{\Delta K}{\frac{1}{2} I_1 \omega_1^2} \]

### Step 8: Final calculation
Substituting the values:
\[ \frac{\Delta K}{K_1} = -0.03 \]
Thus, the percentage change in kinetic energy is:
\[ \text{Percentage change} = -0.03 \times 100\% = -3\% \]

### Final Answer
The percentage change in the rotational kinetic energy is **-3%**.

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