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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 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.

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 the lowest position? Answer in units of m/s.

Learn how to find the linear speed of the center of a solid disk (6.5 m radius, 31.1 kg mass) released from rest, swinging about a pivot at its rim under gravity (9.8 m/s²). This physics problem involves energy conservation, rotational motion, and moment of inertia concepts, suitable for college-level students.

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