A 2.0 m long pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom
The Correct Answer and Explanation is :
To find the speed of the pendulum bob at the bottom of its swing, we can use the principle of conservation of mechanical energy. The pendulum consists of two forms of energy: potential energy (PE) at the start and kinetic energy (KE) at the bottom of the swing.
- Initial Potential Energy (PE_initial): When the pendulum is released from an angle of 25 degrees, it has gravitational potential energy due to its height above the lowest point. The potential energy can be calculated using the formula:
[
PE = mgh
]
where (h) is the height relative to the lowest point, (m) is the mass of the pendulum bob, and (g) is the acceleration due to gravity (approximately (9.81 \, \text{m/s}^2)). - Finding the height (h): The height (h) can be found using trigonometry. The vertical component of the length of the pendulum is given by:
[
h = L – L \cos(\theta) = L(1 – \cos(\theta))
]
where (L) is the length of the pendulum (2.0 m) and (\theta) is the angle (25 degrees). Thus,
[
h = 2.0 \, \text{m} \times (1 – \cos(25^\circ)) \approx 2.0 \, \text{m} \times (1 – 0.9063) \approx 2.0 \, \text{m} \times 0.0937 \approx 0.3748 \, \text{m}
] - Calculating the potential energy: Substituting (h) into the potential energy formula:
[
PE_{\text{initial}} = mg(0.3748)
] - Final Kinetic Energy (KE_final): At the bottom of the swing, all potential energy has been converted into kinetic energy:
[
KE = \frac{1}{2} mv^2
]
By conservation of energy:
[
PE_{\text{initial}} = KE_{\text{final}} \Rightarrow mg(0.3748) = \frac{1}{2} mv^2
]
The mass (m) cancels out:
[
g(0.3748) = \frac{1}{2} v^2 \Rightarrow v^2 = 2g(0.3748) \Rightarrow v^2 = 2 \cdot 9.81 \cdot 0.3748
]
[
v^2 \approx 7.355 \Rightarrow v \approx \sqrt{7.355} \approx 2.71 \, \text{m/s}
]
Therefore, the speed of the bob at the bottom of the swing is approximately 2.71 m/s.