Question

A pendulum bob has a speed of 4 m/s at its lowest position. The pendulum is 1 m long. When the length of the string makes an angle of 60° with the vertical, the speed of the bob at that position is (acceleration due to gravity, \( g = 10 \, \text{m/s}^2, \cos(60^\circ) = 0.5) \)

• A. 6 m/s
• B. \( \sqrt{3} \, \text{m/s} \)
• C. \( \sqrt{6} \, \text{m/s} \)
• D. 3 m/s

Hint:

The speed of a pendulum bob at any position can be found using conservation of mechanical energy. The sum of kinetic and potential energies is constant throughout the motion.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We use the Principle of Conservation of Mechanical Energy. Kinetic energy is converted into potential energy as the bob rises.
Step 2: Key Formula or Approach:
1. Height \( h = L(1 - \cos \theta) \).
2. \( \frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + mgh \).
Step 3: Detailed Explanation:
Given: \( v_0 = 4\text{ m/s} \), \( L = 1\text{ m} \), \( \theta = 60^\circ \).
Calculate height \( h \):
\[ h = 1(1 - \cos 60^\circ) = 1(1 - 0.5) = 0.5\text{ m} \].
Apply energy conservation:
\[ v^2 = v_0^2 - 2gh \]
\[ v^2 = (4)^2 - 2(10)(0.5) \]
\[ v^2 = 16 - 10 = 6 \]
\[ v = \sqrt{6}\text{ m/s} \]
Step 4: Final Answer:
The speed at that position is \( \sqrt{6}\text{ m/s} \).