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 solve the problem, we need to find the angular acceleration of a hollow sphere when a tangential force is applied. We will follow these steps:

### Step 1: Identify the given values
- Mass of the hollow sphere (m) = 1 kg
- Radius of the hollow sphere (R) = 10 cm = 0.1 m
- Tangential force (F) = 30 N

### Step 2: Calculate the moment of inertia (I)
For a hollow sphere rotating about its diameter, the moment of inertia is given by the formula:
\[ I = \frac{2}{3} m R^2 \]
Substituting the values:
\[ I = \frac{2}{3} \times 1 \, \text{kg} \times (0.1 \, \text{m})^2 \]
\[ I = \frac{2}{3} \times 1 \times 0.01 = \frac{2}{300} = \frac{1}{150} \, \text{kg m}^2 \]

### Step 3: Calculate the torque (τ)
The torque (τ) produced by the tangential force is given by:
\[ \tau = R \times F \]
Substituting the values:
\[ \tau = 0.1 \, \text{m} \times 30 \, \text{N} = 3 \, \text{N m} \]

### Step 4: Relate torque to angular acceleration (α)
Using the relation between torque and angular acceleration:
\[ \tau = I \alpha \]
We can rearrange this to find angular acceleration (α):
\[ \alpha = \frac{\tau}{I} \]

### Step 5: Substitute the values to find α
Substituting the values of τ and I:
\[ \alpha = \frac{3 \, \text{N m}}{\frac{1}{150} \, \text{kg m}^2} \]
\[ \alpha = 3 \times 150 = 450 \, \text{rad/s}^2 \]

### Final Answer
The angular acceleration of the hollow sphere is:
\[ \alpha = 450 \, \text{rad/s}^2 \]

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