A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric s

Q: Wait, why is the radius of the path $R + \frac{d}{2}$ and not just $R$?

Because we must track the *center of mass* of the ball. The inner sphere has radius $R$, and the gap is $d$. The ball fits snugly in this gap, so its diameter is approximately $d$, meaning its radius is $\frac{d}{2}$. Therefore, its center is at a distance of $R + \frac{d}{2}$ from the center of the spheres.

Q: Why do we assume the normal reaction $N$ is radially outwards initially?

At the very top, the ball is resting on the inner sphere, so the inner sphere pushes it outwards. We set up our equation assuming this outward push. If our calculated $N$ turns out positive, our assumption was right. If it turns out negative, it means the force is actually acting inwards (provided by the outer sphere).

Q: Why does the ball lose contact with the inner sphere at all?

To move in a circle, the ball needs a specific amount of centripetal force ($\frac{mv^2}{r}$). As it falls, its speed $v$ increases, so it needs *more* centripetal force. However, the inward component of gravity ($mg\cos\theta$) *decreases* as $\theta$ increases. Eventually, gravity alone isn't enough to provide the required centripetal force, so the ball tries to fly outward, losing contact with the inner sphere.

Q: Why is $N_B$ the negative of $N_A$?

Our initial equation $mg\cos\theta - N = \frac{mv^2}{r}$ defined $N$ as an *outward* force. When $N$ becomes negative, it mathematically represents an *inward* force. Since sphere $B$ is on the outside, it can only push inwards. Therefore, the magnitude of the force from sphere $B$ is simply the absolute value of that negative $N$.

Q: What happens at the bottom-most point ($\theta = \pi$)?

At the bottom, $\cos\pi = -1$. The ball is moving at its maximum speed. Gravity is pulling straight down (outwards from the center), and the required centripetal force is huge. The outer sphere $B$ must push inwards with a massive force to overcome gravity and provide the centripetal acceleration. Plugging $\cos\theta = -1$ into our formula gives $N_B = 5mg$.

Question

A spherical ball of mass $m$ is kept at the highest point in the space between two fixed, concentric spheres $A$ and $B$ (see fig.). The smaller sphere $A$ has a radius $R$ and the space between the two spheres has a width $d$. The ball has a diameter very slightly less than $d$. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by $	heta$.
(a) Express the total normal reaction force exerted by the spheres on the ball as a function of angle $	heta$.
(b) Let $N_A$ and $N_B$ denote the magnitudes of the normal reaction forces on the ball exerted by the spheres $A$ and $B$, respectively. Sketch the variations of $N_A$ and $N_B$ as function of $\cos	heta$ in the range $0 \le 	heta \le \pi$ by drawing two separate graphs in your answer book, taking $\cos	heta$ on the horizontal axis.

Sigma Percentile · Accepted Answer

### The Setup: A Constrained Journey

Imagine you are designing a roller coaster, but instead of a track, the cart is trapped in a narrow gap between two massive, fixed concentric spheres. This is exactly the physical reality of our problem. We have a small ball of mass $m$ resting at the very top of the gap between an inner sphere $A$ (radius $R$) and an outer sphere $B$. The gap has a width $d$, and the ball fits snugly inside it.

Because we must track the physics of the ball's center of mass, we first need to determine its path. The center of the ball is located exactly halfway across the gap. Therefore, it moves in a perfect circle whose radius is $R' = R + \frac{d}{2}$. This geometric constraint is the foundation of everything that follows.

### The Engine of Motion: Energy Conservation

When the ball is given a gentle push, it begins to slide down the frictionless gap. As it descends, it loses gravitational potential energy, which is entirely converted into kinetic energy. Let's freeze time when the ball's radius vector makes an angle $	heta$ with the upward vertical.

How far has the ball fallen vertically? By looking at the geometry, the vertical drop $h$ is simply the difference between the initial height $R'$ and the current vertical height $R'\cos	heta$. Thus, $h = R'(1 - \cos	heta)$.

Using the Conservation of Mechanical Energy (or simply the kinematic relation for a body falling under gravity along a smooth curve), the velocity $v$ of the ball is given by $v^2 = 2gh$. Substituting our height, we get the master velocity equation:

$$v^2 = 2gR'(1 - \cos	heta)$$

### The Tug of War: Forces in Circular Motion

Now, let's zoom in on the forces acting on the ball at this angle $	heta$. There are only two players in this game: gravity and the normal reaction from the spheres. Gravity pulls straight down with a force $mg$. However, for circular motion, we only care about the forces acting along the radial line (towards or away from the center).

The component of gravity pulling the ball *towards* the center is $mg\cos	heta$.

What about the normal reaction? At the beginning of the journey, the ball is resting on the inner sphere $A$. So, let's assume the inner sphere is pushing the ball *outwards* with a normal force $N$.

According to Newton's Second Law for circular motion, the net force towards the center must equal the required centripetal force. Therefore, the inward gravity minus the outward normal push gives us our radial equation:

$$mg\cos	heta - N = \frac{mv^2}{R'}$$

### The Critical Moment: Losing Contact

Let's substitute our velocity equation into this force equation. The radius $R'$ beautifully cancels out, leaving us with:

$$mg\cos	heta - N = 2mg(1 - \cos	heta)$$

Rearranging this to solve for the normal reaction $N$, we find:

$$N = mg(3\cos	heta - 2)$$

This equation tells a fascinating story. As the ball falls, $	heta$ increases, which means $\cos	heta$ decreases. Consequently, the outward push $N$ required from the inner sphere is steadily dropping.

Why is this happening? Because as the ball speeds up, it needs *more* centripetal force to stay in the circle. But at the same time, the inward component of gravity ($mg\cos	heta$) is getting *weaker*. The ball is desperately trying to fly outward in a straight line, and gravity is losing its grip.

Eventually, $N$ drops to exactly zero. This happens when $3\cos	heta - 2 = 0$, or $\cos	heta = \frac{2}{3}$. At this precise angle, gravity provides exactly the right amount of centripetal force. The inner sphere doesn't need to push anymore. The ball is momentarily in free fall along the circular path!

### The Handoff: The Outer Sphere Takes Over

What happens if the ball continues past this critical angle? For $\cos	heta < \frac{2}{3}$, our equation yields a *negative* value for $N$.

In physics, a negative sign on a vector simply means it points in the opposite direction. A negative outward force means we need an *inward* force. But the inner sphere $A$ can only push outwards; it cannot pull the ball in!

If the outer sphere $B$ weren't there, the ball would fly off the inner sphere and follow a parabolic projectile path. But sphere $B$ *is* there, acting as a ceiling. It steps in and provides the necessary inward push to keep the ball in the circular gap.

Therefore, the normal reaction from the outer sphere, $N_B$, is simply the magnitude of this required inward force. It is the exact negative of our original $N$:

$$N_B = -N = mg(2 - 3\cos	heta)$$

### Visualizing the Forces: The Graphs

To truly master this concept, we must visualize it. We are asked to plot $N_A$ and $N_B$ against $\cos	heta$.

For the inner sphere $A$, the force $N_A = mg(3\cos	heta - 2)$ is valid only while the ball is in contact with it, which is the domain $\cos	heta \in [\frac{2}{3}, 1]$. This is a simple straight line. At the very top ($\cos	heta = 1$), $N_A = mg$. It linearly drops to zero at $\cos	heta = \frac{2}{3}$.

For the outer sphere $B$, the force $N_B = mg(2 - 3\cos	heta)$ takes over for the rest of the journey, in the domain $\cos	heta \in [-1, \frac{2}{3}]$. This is also a straight line. It starts at zero when $\cos	heta = \frac{2}{3}$. As the ball reaches the absolute bottom of the spheres ($\cos	heta = -1$), the required inward push becomes massive. Plugging in $-1$, we get $N_B = mg(2 - 3(-1)) = 5mg$.

At the bottom, the outer sphere must support the entire weight of the ball *plus* provide the immense centripetal force required for its maximum speed. It's a beautiful, seamless handoff of forces, dictated entirely by the elegant laws of circular motion.

Visualized Solution

Geometry of the Path

Conservation of Energy

Vertical Drop

Velocity at Angle $θ$

Free Body Diagram

Radial Force Equation

Substituting Velocity

Simplifying the Equation

Total Normal Reaction

Condition for Losing Contact

Normal Reaction $N_{A}$

Contact with Outer Sphere

Normal Reaction $N_{B}$

Graph of $N_{A}$ vs $cos θ$

Graph of $N_{B}$ vs $cos θ$

The Sigma Insight: Vertical Circular Motion

The Setup

The Engine of Motion

The Tug of War

The Critical Moment

The Handoff

Visualizing the Forces

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