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A uniform solid disk of radius 6.5 m and
mass 31.1 kg is free to rotate on a frictionless
pivot through a point on its rim.
The acceleration of gravity is 9.8 m/s^2. Initially, the disk is held with its center at
pivot’s height — see the blue circle on the
above diagram. The disk is released and it
swings back and forth on the pivot. On the
diagram, the gray circle with a dashed border shows the lowest position of the disk, when
its center is directly below the pivot. What is
the linear speed of the disk’s center when it
reaches this position?
Answer in units of m/s
.

To find the linear speed of the disk's center when it reaches its lowest position, we can use the principle of conservation of energy. Here’s how we proceed:

### Step-by-Step Solution

1. **Define the Initial and Final Energy States**:
   - **Initial Position**: The disk starts at a height above the lowest point. The potential energy at the initial position is maximum, and the kinetic energy is zero (since it’s initially at rest).
   - **Final Position**: When the disk’s center reaches the lowest point, all the potential energy will have converted into kinetic energy (both translational and rotational) as the disk swings down.

2. **Calculate the Initial Potential Energy**:
   - The center of mass of the disk drops by a vertical distance equal to the radius, \( R = 6.5 \, \text{m} \).
   - The initial gravitational potential energy \( U_i \) (relative to the lowest position) is:
     \[
     U_i = m \cdot g \cdot h = (31.1 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) \cdot (6.5 \, \text{m}) = 1976.7 \, \text{J}
     \]

3. **Apply Conservation of Energy**:
   - At the lowest point, this potential energy is converted into the kinetic energy of the disk’s center (translational) and the rotational kinetic energy about its center.
   - **Total Kinetic Energy** \( K \) at the lowest point is:
     \[
     K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2
     \]
     where:
     - \( v \) is the linear speed of the disk's center,
     - \( I \) is the moment of inertia of the disk about its center,
     - \( \omega \) is the angular speed.

4. **Relate Linear and Angular Speed**:
   - Since the disk rotates about its pivot point, we have \( v = R \omega \).

5. **Moment of Inertia**:
   - The moment of inertia of a disk about its center is \( I_{\text{center}} = \frac{1}{2} m R^2 \).
   - Using the parallel axis theorem to find the moment of inertia about the pivot point:
     \[
     I_{\text{pivot}} = I_{\text{center}} + m R^2 = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2
     \]

6. **Rewrite Kinetic Energy with Known Quantities**:
   - Substitute \( I = \frac{3}{2} m R^2 \) and \( v = R \omega \) into the kinetic energy equation:
     \[
     K = \frac{1}{2} m v^2 + \frac{1}{2} \cdot \frac{3}{2} m R^2 \cdot \left(\frac{v}{R}\right)^2
     \]
   - Simplify:
     \[
     K = \frac{1}{2} m v^2 + \frac{3}{4} m v^2 = \frac{5}{4} m v^2
     \]

7. **Set Up Conservation of Energy Equation**:
   \[
   U_i = K \Rightarrow 1976.7 = \frac{5}{4} m v^2
   \]

8. **Solve for \( v \)**:
   \[
   v^2 = \frac{1976.7 \cdot 4}{5 \cdot m} = \frac{7906.8}{155.5} = 50.85
   \]
   \[
   v = \sqrt{50.85} \approx 7.13 \, \text{m/s}
   \]

### Answer
The linear speed of the disk’s center when it reaches the lowest position is approximately:
\[
v \approx 7.13 \, \text{m/s}
\]

Would you like further details, or do you have any questions?

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Here are 5 related questions for practice:

1. What would happen to the speed if the mass of the disk were doubled?
2. How would the speed change if the pivot point were not at the edge but at the center of the disk?
3. What is the significance of the parallel axis theorem in this calculation?
4. How would air resistance affect the final speed of the disk’s center?
5. If the disk were not uniform, how would the moment of inertia calculation change?

**Tip:** When dealing with rotating bodies, always relate linear and angular speeds using the radius of rotation.

A uniform solid disk of radius 6.5 m and mass 31.1 kg is free to rotate on a frictionless pivot through a point on its rim. The acceleration of gravity is 9.8 m/s^2. Initially, the disk is held with its center at pivot’s height. The disk is released and it swings back and forth on the pivot. What is the linear speed of the disk’s center when it reaches its lowest position? Answer in units of m/s.

Learn how to determine the linear speed of a rotating disk's center when it reaches the lowest position. This physics problem involves principles of energy conservation, rotational kinematics, and the parallel axis theorem. Suitable for physics and engineering students.

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