Question

The string of a pendulum is attached to a bob and is released from its initial rest position. A fixed peg is present in the path of motion of the pendulum. What are the velocities of the bob at positions $B$ and $C$ respectively? The distances are as shown in the figure.

• A. $3.16\text{m/s}$ and $7.07\text{m/s}$
• B. $7.07\text{m/s}$ and $3.16\text{m/s}$
• C. $5.34\text{m/s}$ and $10.78\text{m/s}$
• D. $7.07\text{m/s}$ and $6.32\text{m/s}$

Answer 1

The correct option is D $7.07\text{m/s}$ and $6.32\text{m/s}$

As the bob is released, it swings downwards, Hence its potential energy is converted completely into kinetic energy at point $B$. Taking reference level at point $B$

$K{E}_B=P{E}_A\Rightarrow \frac{1}{2}m{v_B}^2=mgh\Rightarrow {v_B}^2=2gh\Rightarrow v_B=\sqrt{2gh}=\sqrt{2\times 10\times 2.5}\Rightarrow v_B=7.07m/s$
Therefore velocity of the bob at point $B$ is $7.07\text{m/s}$.

From the point $B$ to $C$, only the thread below the peg can swing. Therefore, applying conservation of energy at points $B$ and $C$

$K{E}_C+P{E}_C=K{E}_B=P{E}_A\frac{1}{2}m{v_C}^2+mg{h}_C=mgh\frac{1}{2}m{v_C}^2=mg(h-h_C)\Rightarrow v_C=\sqrt{2g(h-h_C)}\Rightarrow v_C=6.32m/s$