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.