Animated Solution for Physics - Laws of Motion: A small block is shot into each of the four smooth tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in (2001)

Have you ever watched a roller coaster speed into a loop-de-loop and wondered about the intense forces the riders feel at the very top? This classic physics problem captures that exact thrill. We are presented with a small block shot into four different smooth tracks. Each track rises to the exact same maximum height, and the block enters each with the identical initial speed. Our mission is to determine which track exerts the maximum normal reaction on the block at its highest point.

To crack this, we need to blend two powerful pillars of physics: the conservation of mechanical energy and the dynamics of circular motion. Let's break it down step by step.

Speed at the Peak

The first crucial detail in the problem is the word smooth. This tells us that there is absolutely no friction to drain energy from our system. Therefore, the total mechanical energy of the block is perfectly conserved throughout its journey.

Let's compare the energy at the bottom of the track to the energy at the highest point. At the bottom, the block has purely kinetic energy, given by $\frac{1}{2}m{v}^2$. As it climbs, it gains gravitational potential energy, $mgh$, and its speed drops to some value $v_0$ at the top.

By the principle of conservation of mechanical energy, we can write:

Here is the beautiful part: the problem states that the initial speed $v$ and the maximum height $h$ are identical for all four tracks. Since the mass $m$ is also constant, the speed at the highest point, $v_0$, must be exactly the same for every single track!

The Centripetal Demand

Now that we know the block reaches the top of every track with the same speed $v_0$, let's analyze the forces acting on it at that exact moment. Imagine the block upside down at the peak of the curve.

There are two vertical forces acting on the block:

1. Gravity ($mg$): The Earth is pulling the block straight down. 2. Normal Reaction ($N$): The track is physically above the block, so it pushes the block downwards to keep it confined to the circular path.

Because the block is moving along a curved path, it is undergoing circular motion. According to Newton's Second Law, the net force pointing towards the center of the curve must provide the necessary centripetal force.

We can write the master equation for the dynamics at the top:

Where $R$ is the radius of curvature of the track at that specific highest point.

Maximizing the Push

Our goal is to find where the normal reaction $N$ is maximum. Let's rearrange our master equation to isolate $N$:

Take a close look at this equation. The mass $m$, the speed at the top $v_0$, and the acceleration due to gravity $g$ are all constant across the four tracks. The only variable that changes from track to track is $R$, the radius of curvature.

To make the normal reaction $N$ as large as possible, we need the positive term $\frac{m{v}_0^2}{R}$ to be as large as possible. Mathematically, since $R$ is in the denominator, we must minimize $R$.

What does a minimum radius of curvature look like physically? It means the curve is extremely sharp and tight. A larger radius of curvature would look like a gentle, sweeping curve.

When we visually inspect the four tracks provided in the problem, track (a) clearly has the sharpest, tightest curve at its peak. Therefore, track (a) has the minimum radius of curvature $R_{\text{min}}$.

Because it forces the block to make such a tight turn at speed $v_0$, the track must exert a massive downward push to provide the required centripetal force. Thus, the normal reaction is maximum in track (a).