A diver having a moment of inertia of `6.0 kg-m^2` about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5.0 kg-m^2` what will be the new angular speed?

Explanation

To solve the problem, we need to apply the principle of conservation of angular momentum. The angular momentum of a system remains constant if no external torque acts on it.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Moment of inertia (I_initial) = 6.0 kg-m²
- Angular speed (ω_initial) = 2 rad/s

2. **Calculate Initial Angular Momentum:**
- Angular momentum (L_initial) can be calculated using the formula:
\[ L_{\text{initial}} = I_{\text{initial}} \times \omega_{\text{initial}} \]
- Substituting the values:
\[ L_{\text{initial}} = 6.0 \, \text{kg-m}^2 \times 2 \, \text{rad/s} = 12.0 \, \text{kg-m}^2/\text{s} \]

3. **Identify Final Conditions:**
- New moment of inertia (I_final) = 5.0 kg-m²
- We need to find the new angular speed (ω_final).

4. **Apply Conservation of Angular Momentum:**
- Since angular momentum is conserved, we have:
\[ L_{\text{initial}} = L_{\text{final}} \]
- This can be expressed as:
\[ I_{\text{initial}} \times \omega_{\text{initial}} = I_{\text{final}} \times \omega_{\text{final}} \]

5. **Rearranging for ω_final:**
- We can rearrange the equation to solve for ω_final:
\[ \omega_{\text{final}} = \frac{I_{\text{initial}} \times \omega_{\text{initial}}}{I_{\text{final}}} \]

6. **Substituting the Known Values:**
- Substitute I_initial, ω_initial, and I_final into the equation:
\[ \omega_{\text{final}} = \frac{6.0 \, \text{kg-m}^2 \times 2 \, \text{rad/s}}{5.0 \, \text{kg-m}^2} \]
- Calculate:
\[ \omega_{\text{final}} = \frac{12.0 \, \text{kg-m}^2/\text{s}}{5.0 \, \text{kg-m}^2} = 2.4 \, \text{rad/s} \]

7. **Final Answer:**
- The new angular speed (ω_final) is **2.4 rad/s**.