Animated Solution for Physics - Work, Energy, and Power: A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$. The magnitude of the change in its velocity as it reaches a position, where the string is horizontal, is

The problem of a mass whirling in a vertical circle is a classic staple of physics, blending the elegance of kinematics with the raw power of energy conservation. But this particular question hides a subtle trap—one that catches many students off guard. It doesn't just ask for the speed at a new position; it asks for the magnitude of the change in velocity.

Let's embark on this journey and unravel the physics step by step.

Imagine a stone of mass $m$ tied to a string of length $L$. It is launched from the lowest point of its circular path with an initial speed $u$. Because it is constrained to move in a circle, its velocity vector at this lowest point, let's call it $\vec{u}$, is perfectly horizontal and tangent to the path.

As the stone swings upward, gravity pulls it down, slowing it down. Eventually, it reaches a position where the string is perfectly horizontal. At this exact moment, the tangent to the circular path is vertical. Therefore, its new velocity vector, $\vec{v}$, points straight up.

Notice the profound shift: the stone hasn't just lost speed; its direction of motion has rotated by exactly 90 degrees.

Energy Conservation

To find out exactly how much speed the stone has lost, we turn to the principle of Conservation of Mechanical Energy. As the stone moves, two forces act on it: gravity and the tension in the string. However, tension always acts radially inward, perpendicular to the instantaneous velocity. Thus, tension does zero work. Gravity is the only force doing work, meaning mechanical energy is conserved.

Let's set our reference level for gravitational potential energy at the lowest point, so $U=0$ there.

At the lowest point, the total energy is purely kinetic:

When the stone reaches the horizontal position, it has climbed a vertical height equal to the length of the string, $L$. Its total energy is now a mix of kinetic and potential:

Equating the initial and final energies gives us our master equation:

We can cancel the mass $m$ from all terms and multiply the entire equation by 2 to clear the fractions:

Rearranging this to solve for the square of the final speed, we get:

This equation tells us exactly how much the speed has decreased due to the climb against gravity.

Here is where the trap springs. A hasty student might simply calculate the difference in speeds, $v-u$, and look for that in the options. But the question explicitly asks for the change in velocity.

Velocity is a vector. It has both magnitude and direction. The change in velocity is defined as the vector difference:

To visualize this, imagine drawing both velocity vectors from a common origin. The initial velocity $\vec{u}$ points horizontally to the right. The final velocity $\vec{v}$ points vertically upwards. The vector $\Delta \vec{v}$ is the arrow drawn from the tip of $\vec{u}$ to the tip of $\vec{v}$.

Because the initial and final velocities are perpendicular to each other, they form the legs of a right-angled triangle, with $\Delta \vec{v}$ as the hypotenuse.

To find the magnitude of this change in velocity, $|\Delta \vec{v}|$, we simply apply the Pythagorean theorem to our vector triangle:

This is a beautiful geometric result. Now, we bring back the physics. We substitute our energy conservation result, $v^2=u^2-2gL$, into this geometric equation:

Combining the $u^2$ terms, we get:

Finally, factoring out the 2 yields our elegant final answer:

By respecting the vector nature of velocity and combining it with the scalar power of energy conservation, we've safely navigated the trap and arrived at the exact solution.