Two thin rectangular sheets `(0.20 m xx 0.40 m)` are identical. In the first sheet the axis of rotation lies along the 0.20-m side, and in the second it lies along the 0.40-m side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 8.0 s. How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

Options

A. 0.86 s
B. 1.4 s
C. 2.0 s
D. 1.1 s

Answer

To solve the problem, we need to analyze the two sheets and their moments of inertia, as well as the relationship between torque, angular acceleration, and time.

### Step-by-step Solution:

1. **Identify the Dimensions and Setup**:
- We have two identical rectangular sheets with dimensions \(0.20 \, \text{m} \times 0.40 \, \text{m}\).
- For the first sheet, the axis of rotation is along the \(0.20 \, \text{m}\) side.
- For the second sheet, the axis of rotation is along the \(0.40 \, \text{m}\) side.

2. **Calculate the Moment of Inertia**:
- The moment of inertia \(I\) for a rectangular sheet about an axis along one of its sides is given by:
\[ I = \frac{1}{3} m L^2 \]
- For the first sheet (axis along \(0.20 \, \text{m}\)):
\[ I_1 = \frac{1}{3} m (0.40)^2 = \frac{1}{3} m (0.16) = \frac{0.16m}{3} \]
- For the second sheet (axis along \(0.40 \, \text{m}\)):
\[ I_2 = \frac{1}{3} m (0.20)^2 = \frac{1}{3} m (0.04) = \frac{0.04m}{3} \]

3. **Relate Torque, Angular Acceleration, and Time**:
- The torque \(\tau\) applied to each sheet is related to the moment of inertia and angular acceleration \(\alpha\) by:
\[ \tau = I \alpha \]
- For the first sheet:
\[ \tau = I_1 \alpha_1 \implies \tau = \frac{0.16m}{3} \alpha_1 \]
- For the second sheet:
\[ \tau = I_2 \alpha_2 \implies \tau = \frac{0.04m}{3} \alpha_2 \]

4. **Set Up the Ratio of Angular Accelerations**:
- Since the same torque is applied to both sheets:
\[ \frac{0.16m}{3} \alpha_1 = \frac{0.04m}{3} \alpha_2 \]
- Canceling \(m\) and \(\frac{1}{3}\) gives:
\[ 0.16 \alpha_1 = 0.04 \alpha_2 \implies \frac{\alpha_2}{\alpha_1} = \frac{0.16}{0.04} = 4 \implies \alpha_2 = 4 \alpha_1 \]

5. **Relate Time and Angular Velocity**:
- The angular velocity \(\omega\) after time \(t\) is given by:
\[ \omega = \alpha t \]
- For the first sheet:
\[ \omega = \alpha_1 t_1 \]
- For the second sheet:
\[ \omega = \alpha_2 t_2 \]
- Setting the two equations for \(\omega\) equal gives:
\[ \alpha_1 t_1 = \alpha_2 t_2 \implies \alpha_1 t_1 = 4 \alpha_1 t_2 \]
- Dividing both sides by \(\alpha_1\) (assuming \(\alpha_1 \neq 0\)):
\[ t_1 = 4 t_2 \]

6. **Substituting Known Values**:
- We know \(t_1 = 8.0 \, \text{s}\):
\[ 8.0 = 4 t_2 \implies t_2 = \frac{8.0}{4} = 2.0 \, \text{s} \]

### Final Answer:
The time it takes for the second sheet to reach the same angular velocity is **2.0 seconds**.

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