A thin, cylindrical rod 36 cm long has mass 1.91 kg and radius 1.04 cm. A 29.2 kg ball of diameter 8.03 cm is attached to one end.

The arrangement is originally vertical with the ball at the top and is free to pivot about the other end.

8.03 cm

C

kg

1.91

90°

29.2 kg

pivot .

1.91 kg

O

• - 36 cm - »

After the ball-rod system falls a quarter turn, what is its rotational kinetic energy?

The acceleration of gravity is 9.8 m/s?.

Answer in units of J. Answer in units of J.

part 2 of 3

At the same point, what is its angular speed?

of rad/s.

Answer in units of rads. Answer in units

part 3 of 3

At the same point, what is the linear speed of the ball?

ms.

Answer in units of m/s. Answer in units of

To solve this problem, we need to analyze the system as it pivots from the vertical position to a quarter turn (90 degrees). We'll use the principles of rotational dynamics and energy conservation.

### Part 1: Rotational Kinetic Energy (KE)

1. **Calculate the Moment of Inertia (I)**:
   - For the rod (cylinder), the moment of inertia about one end is given by:
     \[
     I_{\text{rod}} = \frac{1}{3} m_{\text{rod}} L^2
     \]
     where \( m_{\text{rod}} = 1.91 \, \text{kg} \) and \( L = 0.36 \, \text{m} \).

   - For the ball (treated as a point mass at distance \( r \) from the pivot):
     \[
     I_{\text{ball}} = m_{\text{ball}} r^2
     \]
     where \( m_{\text{ball}} = 29.2 \, \text{kg} \) and \( r = 0.36 \, \text{m} + 0.04 \, \text{m} = 0.40 \, \text{m} \) (adding the radius of the ball).

2. **Calculate I for both components**:
   - For the rod:
     \[
     I_{\text{rod}} = \frac{1}{3} (1.91) (0.36)^2 = \frac{1.91 \cdot 0.1296}{3} \approx 0.082 \, \text{kg m}^2
     \]
   - For the ball:
     \[
     I_{\text{ball}} = 29.2 \cdot (0.40)^2 = 29.2 \cdot 0.16 = 4.672 \, \text{kg m}^2
     \]

3. **Total Moment of Inertia (I_total)**:
   \[
   I_{\text{total}} = I_{\text{rod}} + I_{\text{ball}} \approx 0.082 + 4.672 \approx 4.754 \, \text{kg m}^2
   \]

4. **Calculate the gravitational potential energy (PE)** before the fall:
   - The height of the center of mass of the ball when vertical is:
     \[
     h_{\text{ball}} = 0.40 \, \text{m}
     \]
   - PE at the top:
     \[
     PE = m_{\text{ball}} g h = 29.2 \cdot 9.8 \cdot 0.40 \approx 114.56 \, \text{J}
     \]

5. **At the bottom (after falling 90 degrees)**, all potential energy converts to kinetic energy:
   \[
   KE = PE \approx 114.56 \, \text{J}
   \]

### Part 2: Angular Speed (ω)

1. **Use the relationship**:
   \[
   KE = \frac{1}{2} I_{\text{total}} \omega^2
   \]

2. **Solve for ω**:
   \[
   114.56 = \frac{1}{2} \cdot 4.754 \cdot \omega^2
   \]
   \[
   \omega^2 = \frac{229.12}{4.754} \approx 48.21
   \]
   \[
   \omega \approx \sqrt{48.21} \approx 6.95 \, \text{rad/s}
   \]

### Part 3: Linear Speed (v) of the Ball

1. **Use the relation**:
   \[
   v = r \cdot \omega
   \]
   where \( r = 0.40 \, \text{m} \):
   \[
   v = 0.40 \cdot 6.95 \approx 2.78 \, \text{m/s}
   \]

### Final Answers:

1. **Rotational Kinetic Energy**: \( \approx 114.56 \, \text{J} \)
2. **Angular Speed**: \( \approx 6.95 \, \text{rad/s} \)
3. **Linear Speed of the Ball**: \( \approx 2.78 \, \text{m/s} \)

If you have any questions or need further details, let me know! Here are some related questions to consider:

1. What would happen to the kinetic energy if the mass of the ball increased?
2. How does the radius of the rod affect the moment of inertia?
3. What would be the potential energy at different angles of the pivot?
4. How would friction at the pivot affect the system's energy?
5. Can you derive the angular speed using conservation of energy principles?

A thin, cylindrical rod 36 cm long has mass 1.91 kg and radius 1.04 cm. A 29.2 kg ball of diameter 8.03 cm is attached to one end. After the ball-rod system falls a quarter turn, what is its rotational kinetic energy, angular speed, and linear speed of the ball?

Explore the calculation of rotational kinetic energy, angular speed, and linear speed of a ball attached to a rod after a quarter turn. This problem involves a rod of 1.91 kg and a ball of 29.2 kg, detailing key principles in rotational dynamics and energy conservation.

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