To find the linear speed of the disk's center when it reaches the lowest position, we can use the principle of energy conservation. Here’s a step-by-step solution:

1. Setup

• Mass of disk (m) = $31.1 \, \text{kg}$
• Radius of disk (r) = $6.5 \, \text{m}$
• Gravitational acceleration (g) = $9.8 \, \text{m/s}^2$

When the disk is released from rest, its initial potential energy will be converted to kinetic energy when it swings down to its lowest position.

2. Initial Potential Energy

Initially, the center of mass of the disk is located one radius ($r$) above the lowest position. Therefore, the gravitational potential energy at the starting position relative to the lowest position is given by:

$PE_{\text{initial}} = m \cdot g \cdot h = m \cdot g \cdot r$

Substitute the values:

$PE_{\text{initial}} = 31.1 \cdot 9.8 \cdot 6.5 = 1979.87 \, \text{J}$

3. Kinetic Energy at the Lowest Point

At the lowest position, the disk has both rotational kinetic energy and translational kinetic energy. The total kinetic energy ($KE$) is given by:

$KE = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$

where:

• $I$ is the moment of inertia of the disk about the pivot.
• $\omega$ is the angular speed of the disk.
• $v$ is the linear speed of the center of mass.

4. Moment of Inertia of the Disk about the Pivot

The moment of inertia of a disk about an axis through its rim (using the parallel axis theorem) is:

$I = I_{\text{center}} + m \cdot r^2 = \frac{1}{2} m r^2 + m \cdot r^2 = \frac{3}{2} m r^2$

Substitute the values:

$I = \frac{3}{2} \cdot 31.1 \cdot (6.5)^2 = 1971.075 \, \text{kg} \cdot \text{m}^2$

5. Conservation of Energy

According to energy conservation:

$PE_{\text{initial}} = KE$

Thus,

$1979.87 = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$

Since $v = \omega \cdot r$, we can substitute $\omega = \frac{v}{r}$:

$1979.87 = \frac{1}{2} \cdot 1971.075 \cdot \left(\frac{v}{6.5}\right)^2 + \frac{1}{2} \cdot 31.1 \cdot v^2$

6. Solving for $v$

Rearrange and solve this equation for $v$. Let me calculate this part.The solution for the linear speed $v$ of the disk's center when it reaches the lowest position is approximately:

$v = 7.14 \, \text{m/s}$

Since speed cannot be negative, we take the positive value: $v = 7.14 \, \text{m/s}$.

Would you like more details on any part of this solution, or do you have any further questions?

Here are some related questions to explore further:

1. What would happen if the disk had a different mass?
2. How would the speed change if the pivot was located at a different point on the disk?
3. What if the disk were rotating in a medium with friction?
4. How would we approach this problem if we used torque and angular acceleration?
5. What would be the potential energy and speed if the disk was only partially lifted?

Tip: Always remember to consider both rotational and translational kinetic energy in conservation of energy problems involving rotation.