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: Conservation of mechanical energy.
For the pendulum, we can apply the conservation of mechanical energy. The total mechanical energy (sum of kinetic energy and potential energy) remains constant throughout the motion of the pendulum. The energy at the lowest point (where speed is maximum) is entirely kinetic, while at the position where the string makes an angle \( \theta \) with the vertical, the energy is shared between kinetic and potential energy.

Step 2: Energy at the lowest position.
At the lowest position, all the potential energy has been converted into kinetic energy. The total mechanical energy \( E_{\text{total}} \) at the lowest point is:
\[ E_{\text{total}} = \frac{1}{2} m v_{\text{max}}^2, \]
where \( v_{\text{max}} = 4 \, \text{m/s} \) is the speed at the lowest point.
Thus, the total energy is:
\[ E_{\text{total}} = \frac{1}{2} m (4)^2 = 8 m \, \text{J}. \]

Step 3: Energy at the position where the angle is 60°.
At the position where the angle is 60°, the energy is shared between the kinetic and potential energies. The potential energy at this position is:
\[ U = mgh = mgL(1 - \cos(\theta)), \]
where:
- \( L = 1 \, \text{m} \) is the length of the pendulum,
- \( \theta = 60^\circ \) is the angle between the string and the vertical,
- \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity.
The height \( h \) is given by \( h = L(1 - \cos(\theta)) \), so:
\[ U = mgL(1 - \cos(60^\circ)) = 10 m \times 1 \times (1 - 0.5) = 5 m \, \text{J}. \]

Step 4: Kinetic energy at the same position.
The total mechanical energy at this position is still \( E_{\text{total}} = 8 m \, \text{J} \), so the kinetic energy \( K \) at this position is:
\[ K = E_{\text{total}} - U = 8 m - 5 m = 3 m \, \text{J}. \]
Thus, the speed at this position is found from the kinetic energy:
\[ K = \frac{1}{2} m v^2 \quad \Rightarrow \quad 3 m = \frac{1}{2} m v^2. \]
Solving for \( v \):
\[ v^2 = 6 \quad \Rightarrow \quad v = \sqrt{6} \, \text{m/s}. \]
Final Answer:
Thus, the speed of the bob at that position is:
\[ \boxed{3 \, \text{m/s}}. \]