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?

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.