https://file.aiquickdraw.com/mathbot/file/173082357147115iprgmo.jpg

To solve this problem, let's analyze the information provided and calculate the magnitude of \( F_2 \).

### Given Data:
1. **Radius of disk** (\( R \)) = 1.6 cm = 0.016 m
2. **Mass of disk** (\( m \)) = 20 grams = 0.02 kg
3. **Initial angular velocity** (\( \omega_0 \)) = 0 (since it starts from rest)
4. **Final angular velocity** (\( \omega \)) = 280 rad/s (counterclockwise)
5. **Time interval** (\( t \)) = 1.2 s
6. **Force \( F_1 \)** = 0.096 N

The question requires us to find the magnitude of the force \( F_2 \).

### Approach:
1. **Calculate the moment of inertia (I) of the disk**:
   For a uniform disk rotating around its center, the moment of inertia is:
   \[
   I = \frac{1}{2} m R^2
   \]
   
2. **Calculate the angular acceleration (\( \alpha \))**:
   Using the formula for angular acceleration:
   \[
   \alpha = \frac{\omega - \omega_0}{t}
   \]

3. **Calculate the net torque (\( \tau \))**:
   Torque (\( \tau \)) is related to angular acceleration by:
   \[
   \tau = I \cdot \alpha
   \]

4. **Set up the equation for net torque with \( F_1 \) and \( F_2 \)**:
   Since the forces are applied tangentially, the torque due to each force is:
   \[
   \tau = F_1 \cdot R - F_2 \cdot R
   \]
   Since we know \( \tau \), we can solve for \( F_2 \).

Let's go through the calculations:

1. **Moment of inertia**:
   \[
   I = \frac{1}{2} \times 0.02 \, \text{kg} \times (0.016 \, \text{m})^2 = 2.56 \times 10^{-6} \, \text{kg} \cdot \text{m}^2
   \]

2. **Angular acceleration**:
   \[
   \alpha = \frac{280 \, \text{rad/s} - 0}{1.2 \, \text{s}} = 233.33 \, \text{rad/s}^2
   \]

3. **Net torque**:
   \[
   \tau = I \cdot \alpha = (2.56 \times 10^{-6} \, \text{kg} \cdot \text{m}^2) \times 233.33 \, \text{rad/s}^2 = 0.000597 \, \text{N} \cdot \text{m}
   \]

4. **Torque equation with \( F_1 \) and \( F_2 \)**:
   \[
   \tau = F_1 \cdot R - F_2 \cdot R
   \]
   Substituting \( \tau = 0.000597 \, \text{N} \cdot \text{m} \), \( F_1 = 0.096 \, \text{N} \), and \( R = 0.016 \, \text{m} \):
   \[
   0.000597 = (0.096) \cdot (0.016) - F_2 \cdot (0.016)
   \]

5. **Solve for \( F_2 \)**:
   \[
   F_2 = \frac{(0.096 \cdot 0.016) - 0.000597}{0.016}
   \]
   \[
   F_2 = \frac{0.001536 - 0.000597}{0.016} = \frac{0.000939}{0.016} = 0.0587 \, \text{N}
   \]

### Answer:
The magnitude of \( F_2 \) is approximately \( 0.0587 \, \text{N} \).

Would you like a further breakdown of any part of this solution? 

Here are some related questions you might consider:

1. How would the solution change if the disk had a different mass?
2. What if the disk’s radius were larger? How would \( F_2 \) be affected?
3. How does the direction of the forces impact the angular velocity?
4. Can this approach be applied to non-uniform objects, like a ring?
5. What role does the moment of inertia play in rotational dynamics?

**Tip**: Always double-check unit conversions, especially when switching from grams to kilograms or centimeters to meters.

The figure shows a uniform disk that can rotate around its center like a merry-go-round. The disk has a radius of 1.6 cm and a mass of 20 grams and is initially at rest. Starting at time t = 0, two forces are to be applied tangentially to the rim as indicated, so that at time t = 1.2 s the disk has an angular velocity of 280 rad/s counterclockwise. Force F1 has a magnitude of 0.096 N. What is the magnitude of F2?

This problem explores rotational dynamics by calculating the force F2 required to achieve a specific angular velocity in a disk subjected to two tangential forces. The disk has a radius of 1.6 cm and a mass of 20 grams, with given forces and time. Ideal for students studying rotational motion concepts, including torque and angular acceleration.

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