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To solve this problem, we can use the principle of conservation of angular momentum. Since there is no external torque acting on the system, the angular momentum before and after the student pulls the weights inward must be the same.

### Given Data
- Initial angular speed: \(\omega_i = 3.4 \, \text{rad/s}\)
- Initial moment of inertia: \(I_i = 11.0 \, \text{kg} \cdot \text{m}^2\)
- Final moment of inertia: \(I_f = 5.5 \, \text{kg} \cdot \text{m}^2\)

### 1. Resulting Angular Speed \( \omega_f \)
According to the conservation of angular momentum:
\[
L_i = L_f
\]
where \(L_i = I_i \omega_i\) and \(L_f = I_f \omega_f\).

So,
\[
I_i \omega_i = I_f \omega_f
\]
Rearrange to solve for \(\omega_f\):
\[
\omega_f = \frac{I_i \omega_i}{I_f}
\]

Plugging in the values:
\[
\omega_f = \frac{(11.0 \, \text{kg} \cdot \text{m}^2)(3.4 \, \text{rad/s})}{5.5 \, \text{kg} \cdot \text{m}^2}
\]
\[
\omega_f = \frac{37.4}{5.5} = 6.8 \, \text{rad/s}
\]

So, the resulting angular speed \( \omega_f \) of the platform is **6.8 rad/s**.

### 2. Increase in Kinetic Energy
The rotational kinetic energy \(K\) is given by:
\[
K = \frac{1}{2} I \omega^2
\]

The initial kinetic energy \(K_i\) and the final kinetic energy \(K_f\) are:
\[
K_i = \frac{1}{2} I_i \omega_i^2
\]
\[
K_f = \frac{1}{2} I_f \omega_f^2
\]

Substitute the values to find \(K_i\) and \(K_f\):
\[
K_i = \frac{1}{2} \times 11.0 \, \text{kg} \cdot \text{m}^2 \times (3.4 \, \text{rad/s})^2
\]
\[
K_i = \frac{1}{2} \times 11.0 \times 11.56 = 63.58 \, \text{J}
\]

For \(K_f\):
\[
K_f = \frac{1}{2} \times 5.5 \, \text{kg} \cdot \text{m}^2 \times (6.8 \, \text{rad/s})^2
\]
\[
K_f = \frac{1}{2} \times 5.5 \times 46.24 = 127.16 \, \text{J}
\]

The increase in kinetic energy is:
\[
\Delta K = K_f - K_i
\]
\[
\Delta K = 127.16 \, \text{J} - 63.58 \, \text{J} = 63.58 \, \text{J}
\]

### Answers
1. The resulting angular speed \( \omega_f \) of the platform is **6.8 rad/s**.
2. The increase in kinetic energy of the system is **63.58 J**.

This increase in kinetic energy comes from the work done by the student to pull the weights inward.

Would you like a breakdown of these calculations or have any further questions?

Here are some follow-up questions that expand on this topic:
1. How would the angular speed change if the final moment of inertia were different?
2. Why does pulling the weights inward increase the kinetic energy of the system?
3. What role does torque play in the conservation of angular momentum here?
4. If there were friction on the platform, how would that affect the conservation of angular momentum?
5. How would the results change if the student pushed the weights further outward instead of pulling them inward?

**Tip:** Conservation of angular momentum is a powerful principle in rotational motion, especially useful when dealing with systems where there are no external torques.

A physics student is sitting on a frictionless platform that is rotating with an initial angular speed of ω_i = 3.4 rad/s. His arms are outstretched, holding a heavy weight in each hand. The moment of inertia of the student, the extended weights, and the platform about the rotation axis is I_i = 11.0 kg·m^2. When the student pulls the weights inward, the moment of inertia decreases to I_f = 5.5 kg·m^2. What is the resulting angular speed, ω_f, of the platform? By how much does the kinetic energy of the system increase?

Learn how to solve a physics problem involving the conservation of angular momentum and rotational kinetic energy. A student on a rotating platform with an initial angular speed of 3.4 rad/s pulls weights inward, reducing the system's moment of inertia. This solution details the calculations for the resulting angular speed and the increase in kinetic energy. Suitable for undergraduate students studying physics.

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