To find the speed of the lower end of the rod when it makes an angle of 30 degrees with the horizontal, we can use principles from physics, particularly the conservation of energy and some basic kinematics. Let's break down the problem step by step.
Understanding the Setup
We have a uniform rod of mass \( m = 2 \, \text{kg} \) and length \( l = 7.8 \, \text{m} \) that is initially vertical. When it starts to fall, the lower end remains in contact with the surface, and the rod pivots about this point. As it falls, it will convert potential energy into kinetic energy.
Potential Energy at the Start
Initially, when the rod is vertical, its center of mass is at a height of \( \frac{l}{2} \) from the ground. The potential energy (PE) at this position can be calculated using the formula:
Here, \( h = \frac{l}{2} = \frac{7.8}{2} = 3.9 \, \text{m} \). Thus, the initial potential energy is:
- PE_initial = \( 2 \times 9.81 \times 3.9 \) = \( 76.686 \, \text{J} \)
Potential Energy at 30 Degrees
When the rod makes an angle of 30 degrees with the horizontal, the height of the center of mass changes. The new height \( h' \) can be calculated using trigonometry:
- h' = \( \frac{l}{2} \cos(30^\circ) \)
Calculating this gives:
- h' = \( 3.9 \times \frac{\sqrt{3}}{2} \approx 3.39 \, \text{m} \)
The potential energy at this position is:
- PE_final = \( 2 \times 9.81 \times 3.39 \approx 66.54 \, \text{J} \)
Change in Potential Energy
The change in potential energy as the rod falls from vertical to 30 degrees is:
- ΔPE = PE_initial - PE_final
Calculating this gives:
- ΔPE = \( 76.686 - 66.54 \approx 10.146 \, \text{J} \)
Conservation of Energy
According to the conservation of energy, the loss in potential energy will equal the gain in kinetic energy (KE) of the rod. The kinetic energy of the rod when it is at 30 degrees can be expressed as:
- KE = \( \frac{1}{2} mv^2 \)
Setting the change in potential energy equal to the kinetic energy gives:
- 10.146 = \( \frac{1}{2} \times 2 \times v^2 \)
Solving for \( v \):
- 10.146 = \( v^2 \)
- v = \( \sqrt{10.146} \approx 3.18 \, \text{m/s} \)
Final Result
Therefore, the speed of the lower end of the rod when it makes an angle of 30 degrees with the horizontal is approximately \( 3.18 \, \text{m/s} \). This result illustrates how energy transformations occur in a falling object, showcasing the interplay between potential and kinetic energy in a dynamic system.
