Question

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of $60^\circ$ to the vertical will be ($g = 10 \text{ m/s}^2$) ________ m/s.

Hint:

For \(\theta = 60^\circ\), the height is exactly half the length of the string (\(L/2\)). This is a very common shortcut in physics problems.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The motion of a pendulum is governed by the Law of Conservation of Mechanical Energy. As the bob rises, kinetic energy is converted into potential energy.
Step 2: Key Formula or Approach:
1. Potential energy gain: \(h = L(1 - \cos \theta)\).
2. Energy conservation: \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh\).
Step 3: Detailed Explanation:
Given: \(u = 3 \text{ m/s}\), \(L = 0.5 \text{ m}\), \(\theta = 60^\circ\).
Calculate the vertical height \(h\):
\[ h = 0.5(1 - \cos 60^\circ) = 0.5(1 - 0.5) = 0.25 \text{ m} \]
Using conservation of energy:
\[ \frac{1}{2} (3)^2 = \frac{1}{2} v^2 + (10 \times 0.25) \]
\[ 4.5 = 0.5 v^2 + 2.5 \]
\[ 0.5 v^2 = 2 \implies v^2 = 4 \]
\[ v = 2 \text{ m/s} \]
Step 4: Final Answer:
The speed of the bob at \(60^\circ\) is 2 m/s.