Animated Solution for Physics - Kinematics: In the arrangement shown in the figure, the ends $P$ and $Q$ of an unstretchable string move downwards with uniform speed $U$. Pulleys $A$ and $B$ are fixed. Mass $M$ moves upwards with a speed

Imagine you are looking at a classic pulley system. Two pulleys fixed to a ceiling, a string draped over them, and a mass hanging in the middle. You pull the ends of the string down with a constant speed $U$. Intuition might scream that the mass in the middle should also rise with speed $U$, or maybe $2U$. But physics is rarely about raw intuition; it is about uncovering the hidden constraints that govern motion.

This problem is a beautiful example of constrained motion, where the geometry of the setup strictly dictates how different parts of the system can move relative to each other. Let's embark on a journey to decode this geometry and find the true upward speed of the mass.

The secret to solving this problem lies in drawing a single, imaginary right-angled triangle. Let's visualize a vertical line dropping straight down from the midpoint between the two pulleys, passing right through the center of mass $M$.

Let the horizontal distance from this central vertical line to pulley $B$ be $c$. Because the pulleys are rigidly bolted to the ceiling, this distance $c$ is an absolute constant. It will never change, no matter how much we pull the strings.

Next, let the vertical depth of the mass $M$ from the horizontal line connecting the pulleys be $y$. As the mass moves up, $y$ will decrease. Finally, let the length of the slanted portion of the string, from pulley $B$ to the mass $M$, be $l$.

Look closely at these three lengths: $c$, $y$, and $l$. They form a perfect right-angled triangle! The string length $l$ is the hypotenuse, $c$ is the base, and $y$ is the perpendicular height. Thanks to Pythagoras, we can lock these variables into a master constraint equation:

This equation is the foundational bedrock of our solution. It mathematically binds the length of the string to the vertical position of the mass.

We have an equation for position, but we need to find velocities. In physics, velocity is simply the rate of change of position with respect to time. To transition from static geometry to dynamic motion, we must differentiate our constraint equation with respect to time $t$.

Let's apply the time derivative to both sides:

Using the chain rule, the derivative of $l^2$ becomes $2l\frac{dl}{dt}$. Now, what about $c^2$? Remember our crucial observation: $c$ is a constant. The derivative of any constant is zero. The derivative of $y^2$ becomes $2y\frac{dy}{dt}$.

Putting it all together, we get:

We can immediately simplify this by dividing both sides by 2:

To make the relationship between the rates of change crystal clear, let's isolate $\frac{dy}{dt}$:

Now comes the most critical phase: translating these abstract mathematical derivatives back into the physical velocities given in the problem. This is where many students fall into a trap with sign conventions.

What does $\frac{dl}{dt}$ represent? It is the rate at which the slanted string length $l$ is changing. The problem states that the ends of the string are being pulled downwards with a uniform speed $U$. This means the length $l$ is shrinking at a rate of $U$. Because it is decreasing, the rate of change is negative:

Similarly, what is $\frac{dy}{dt}$? It is the rate at which the vertical depth $y$ is changing. Let the upward speed of the mass $M$ be $v_M$. As the mass moves up, its depth $y$ is also shrinking. Therefore, its rate of change is also negative:

Let's substitute these physical realities back into our simplified calculus equation:

The negative signs on both sides elegantly cancel each other out, leaving us with a clean, positive relationship:

We are almost there! We have the velocity $v_M$ in terms of $U$, but it is multiplied by the ratio of two lengths, $\frac{l}{y}$. The multiple-choice options are given in terms of the angle $\theta$. We need to bridge this final gap using basic trigonometry.

Look back at our right-angled triangle. The problem defines $\theta$ as the angle the string makes with the vertical. In our triangle, the side adjacent to the angle $\theta$ is the vertical depth $y$, and the hypotenuse is the string length $l$.

By the definition of cosine:

Our velocity equation contains the reciprocal of this ratio, $\frac{l}{y}$. Therefore:

For the grand finale, we substitute this trigonometric identity into our velocity equation:

And there we have it! The upward speed of the mass $M$ is exactly $\frac{U}{\cos \theta }$. This perfectly matches option (b).

This problem beautifully demonstrates that in kinematics, you cannot always rely on simple visual intuition. You must anchor your logic in rigid geometric constraints, use calculus to find the rates of change, and carefully apply sign conventions to reveal the true nature of the motion.