Animated Solution for Physics - Work, Energy, and Power: A pendulum bob has a speed of $3\text{ m/s}$ at its lowest position. The pendulum is $50\text{ cm}$ long. The speed of bob when the length makes an angle of $60^{\circ }$ to the vertical will be ....... $\text{m/s}$. (Take, $g=10{\text{ m/s}}^2$)

This problem is a classic application of the Conservation of Mechanical Energy in the context of vertical circular motion. Let's break down the physics step-by-step to understand exactly how the pendulum behaves as it swings upward against gravity.

Imagine a pendulum bob swinging freely. At its lowest point, it possesses its maximum kinetic energy and is moving at a speed of $v_1=3\text{ m/s}$. The length of the pendulum string is $l=50\text{ cm}$, which we must convert to standard SI units: $l=0.5\text{ m}$.

We are tasked with finding the speed of the bob, $v_2$, when the string makes an angle of $\theta =60^{\circ }$ with the vertical. As the bob swings upward, the tension in the string is always perpendicular to the instantaneous velocity of the bob. Because work is the dot product of force and displacement ($W=\vec{F}\cdot d\vec{r}$), the tension does zero work. The only force doing work is gravity, which is a conservative force. This allows us to confidently apply the principle of conservation of mechanical energy.

The total mechanical energy at the lowest point must equal the total mechanical energy at the $60^{\circ }$ position:

To make our calculations elegant, we set the lowest point of the swing as our reference level for potential energy. Therefore, the initial potential energy $P{E}_i=0$.

As the bob swings to an angle $\theta$, it rises by a vertical height $h$. By looking at the geometry of the right triangle formed by the string and the vertical axis, the vertical projection of the string is $l\cos \theta$. The height gained by the bob is the total length minus this projection:

Substituting our energy expressions into the master equation, we get:

Notice a beautiful property of this equation: the mass $m$ appears in every single term. This means we can divide the entire equation by $m$, proving that the mass of the bob has absolutely no effect on its speed at any point in the swing!

Rearranging the equation to solve for $v_2^2$, we multiply by 2 and isolate the final velocity term:

Now, we substitute the known values: $v_1=3\text{ m/s}$, $g=10{\text{ m/s}}^2$, $l=0.5\text{ m}$, and $\theta =60^{\circ }$. Recalling that $\cos 60^{\circ }=\frac{1}{2}$, we proceed with the arithmetic:

Taking the square root gives us the final speed:

As expected, the bob slows down from $3\text{ m/s}$ to $2\text{ m/s}$ as its kinetic energy is converted into gravitational potential energy. This elegant interplay between kinetic and potential energy is the heart of all conservative systems in mechanics.