A hollow sphere of mass 1 kg and radius 10 cm is free to rotate about its diameter. If a force of 30 N is applied tangentially to it, its angular acceleration is (in rad/`s^(2)`)

Options

A. 5000
B. 450
C. 50
D. 5

Answer

To find the angular acceleration of a hollow sphere when a tangential force is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the hollow sphere, \( m = 1 \, \text{kg} \)
- Radius of the hollow sphere, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Tangential force applied, \( F = 30 \, \text{N} \)

2. **Calculate the torque (\( \tau \)) caused by the applied force:**
\[ \tau = r \times F \]
Since the force is applied tangentially, the torque can be calculated as:
\[ \tau = r \cdot F = 0.1 \, \text{m} \cdot 30 \, \text{N} = 3 \, \text{N m} \]

3. **Determine the moment of inertia (\( I \)) of the hollow sphere about its diameter:**
The moment of inertia of a hollow sphere about its diameter is given by the formula:
\[ I = \frac{2}{3} m r^2 \]
Substituting the values:
\[ I = \frac{2}{3} \cdot 1 \, \text{kg} \cdot (0.1 \, \text{m})^2 = \frac{2}{3} \cdot 1 \cdot 0.01 = \frac{2}{300} = \frac{1}{150} \, \text{kg m}^2 \]

4. **Apply Newton's second law for rotation:**
According to Newton's second law for rotation:
\[ \tau = I \alpha \]
Where \( \alpha \) is the angular acceleration. Rearranging the equation gives:
\[ \alpha = \frac{\tau}{I} \]

5. **Substitute the values of torque and moment of inertia into the equation:**
\[ \alpha = \frac{3 \, \text{N m}}{\frac{1}{150} \, \text{kg m}^2} = 3 \cdot 150 = 450 \, \text{rad/s}^2 \]

6. **Final answer:**
The angular acceleration \( \alpha \) is:
\[ \alpha = 450 \, \text{rad/s}^2 \]