A uniform rod of mass 4.2 kg is 8 m long.
The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 8 m from the center of mass of the rod. The rod is released from rest at an initial angle of 58 with respect to the horizontal, as shown.

What is the angular speed of the rod at the instant the rod is in a horizontal position?
The acceleration due to gravity is 9.8 m/s? and the moment of inertia of the rod about its center of mass is Icm= 1/12 ml^2
Answer in units of rad/s. Answer in units of rad/s.

To find the angular speed of the rod when it reaches a horizontal position, we can use the principles of conservation of energy and rotational dynamics.

### Given Data
- Mass of the rod, \( m = 4.2 \) kg
- Length of the rod, \( l = 8 \) m
- Initial angle, \( \theta_0 = 58^\circ \)
- Gravitational acceleration, \( g = 9.8 \) m/s²

### Moment of Inertia
The moment of inertia of the rod about its center of mass is given by:
\[
I_{\text{cm}} = \frac{1}{12} m l^2
\]
Substituting the values:
\[
I_{\text{cm}} = \frac{1}{12} \times 4.2 \, \text{kg} \times (8 \, \text{m})^2 = \frac{1}{12} \times 4.2 \times 64 = 22.4 \, \text{kg} \cdot \text{m}^2
\]

### Parallel Axis Theorem
Since the pivot is not at the center of mass but at a distance \( d = 8 \, \text{m} \) from the center of mass, we apply the parallel axis theorem:
\[
I = I_{\text{cm}} + m d^2
\]
Where \( d = 8 \, \text{m} \):
\[
I = 22.4 + 4.2 \times (8^2) = 22.4 + 4.2 \times 64 = 22.4 + 268.8 = 291.2 \, \text{kg} \cdot \text{m}^2
\]

### Potential Energy Change
When the rod is at the initial angle, it has potential energy given by:
\[
PE = mgh
\]
Where \( h \) is the height of the center of mass relative to the horizontal position. The center of mass is at \( \frac{l}{2} = 4 \, \text{m} \) from the pivot. At an angle \( 58^\circ \):
\[
h = 4 \sin(58^\circ)
\]
Calculating \( h \):
\[
h = 4 \sin(58^\circ) \approx 4 \times 0.5299 \approx 2.1196 \, \text{m}
\]
Thus,
\[
PE = 4.2 \times 9.8 \times 2.1196 \approx 4.2 \times 20.8 \approx 87.696 \, \text{J}
\]

### Kinetic Energy in Horizontal Position
When the rod is horizontal, all the potential energy converts to rotational kinetic energy:
\[
KE = \frac{1}{2} I \omega^2
\]
Setting \( PE = KE \):
\[
87.696 = \frac{1}{2} \times 291.2 \times \omega^2
\]
Solving for \( \omega \):
\[
87.696 = 145.6 \omega^2
\]
\[
\omega^2 = \frac{87.696}{145.6} \approx 0.6014
\]
\[
\omega \approx \sqrt{0.6014} \approx 0.775 \, \text{rad/s}
\]

### Conclusion
The angular speed of the rod when it is in a horizontal position is approximately **0.775 rad/s**.

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. How would the result change if the mass of the rod was different?
2. What would happen if the rod were not uniform?
3. How does the angle of release affect the final angular speed?
4. What is the effect of friction at the pivot point on the result?
5. How would you calculate the linear speed of the end of the rod in the horizontal position?

**Tip:** When analyzing problems involving rotation and energy, always consider the conservation laws, as they simplify the calculations significantly.

A uniform rod of mass 4.2 kg is 8 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 8 m from the center of mass of the rod. The rod is released from rest at an initial angle of 58 degrees with respect to the horizontal. What is the angular speed of the rod at the instant the rod is in a horizontal position?

Explore how to calculate the angular speed of a uniform rod with a mass of 4.2 kg and length of 8 m, pivoted at a distance from its center of mass. This problem incorporates principles of rotational dynamics and conservation of energy, leading to an angular speed of approximately 0.775 rad/s.

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