To find the new angular velocity $\omega'$ when the student is 1.73 m from the center, we can use the principle of conservation of angular momentum. The total angular momentum of the system must remain constant since no external torques are acting on it.

The initial angular momentum $L_i$ when the student is at the rim (3.11 m from the center) can be expressed as:

$L_i = (I_{\text{platform}} + I_{\text{student}}) \omega$

where:

• $I_{\text{platform}} = M r^2$ is the moment of inertia of the platform.
• $I_{\text{student}} = m r^2$ is the moment of inertia of the student when at the rim.

The final angular momentum $L_f$ when the student is 1.73 m from the center is:

$L_f = (I_{\text{platform}} + I'_{\text{student}}) \omega'$

where $I'_{\text{student}} = m (r - d)^2$ and $d = 1.73 \, \text{m}$ is the distance from the center. Therefore, $r - d = 3.11 - 1.73 = 1.38 \, \text{m}$.

Step 1: Calculate Initial Moment of Inertia

1. Moment of inertia of the platform: $I_{\text{platform}} = M r^2 = 135.1 \, \text{kg} \times (3.11 \, \text{m})^2 = 135.1 \times 9.6721 \approx 1305.54 \, \text{kg m}^2$

2. Moment of inertia of the student at the rim: $I_{\text{student}} = m r^2 = 92.3 \, \text{kg} \times (3.11 \, \text{m})^2 = 92.3 \times 9.6721 \approx 891.14 \, \text{kg m}^2$

Step 2: Calculate Initial Angular Momentum

$L_i = (I_{\text{platform}} + I_{\text{student}}) \omega = (1305.54 + 891.14) \times 1.10 \approx 2296.68 \times 1.10 \approx 2526.35 \, \text{kg m}^2/\text{s}$

Step 3: Calculate Final Moment of Inertia

1. Moment of inertia of the student when 1.73 m from the center: $I'_{\text{student}} = m (r - d)^2 = 92.3 \, \text{kg} \times (1.38 \, \text{m})^2 = 92.3 \times 1.9044 \approx 175.64 \, \text{kg m}^2$

2. Total moment of inertia when the student is at 1.73 m: $I_f = I_{\text{platform}} + I'_{\text{student}} = 1305.54 + 175.64 \approx 1481.18 \, \text{kg m}^2$

Step 4: Calculate Final Angular Velocity

Using the conservation of angular momentum: $L_i = L_f \Rightarrow 2526.35 = (1481.18) \omega'$ Solving for $\omega'$: $\omega' = \frac{2526.35}{1481.18} \approx 1.70 \, \text{rad/s}$

Conclusion

The new angular velocity $\omega'$ when the student is 1.73 m from the center is approximately 1.70 rad/s.

If you have any questions or need further details, feel free to ask! Here are some follow-up questions:

1. What is the significance of conservation of angular momentum in this scenario?
2. How would the final angular velocity change if the mass of the student increased?
3. What would happen to the system if the platform had friction?
4. Can you describe how the moment of inertia affects angular velocity?
5. What are real-life applications of these principles in engineering or physics?

Tip: Always check if there are external forces or torques acting on a system before applying conservation laws.