A small sphere of radius R is held against the inner surface of a larger sphere of radius 6R. The ma

Q: Why is the distance between the centers $5R$ and not $6R$?

The small sphere is inside the large sphere. The distance from the center of the large sphere to its inner surface is $6R$. Since the small sphere has a radius of $R$, its center is $R$ distance away from the inner surface. Thus, the distance between their centers is $6R - R = 5R$.

Q: Why doesn't the center of mass move horizontally?

Because all surfaces are frictionless, there is no external horizontal force (like friction) acting on the system. According to Newton's second law, if the net external force is zero, the acceleration of the center of mass is zero, meaning it stays at rest.

Q: Does the center of mass move vertically?

Yes! Gravity acts downwards, and the normal force from the table acts upwards. As the small sphere moves down and up the inner surface, the vertical position of the center of mass will change. However, we only care about the horizontal position for this problem.

Q: Why did we assume the final position of the large sphere is $x$?

Since the small sphere moves to the left, it pushes the large sphere to the right (to conserve the center of mass). We don't know exactly how much it moved yet, so we assign a variable $x$ to its new center and use the conservation principle to solve for it.

Q: Can we solve this using the concept of displacement ($\Delta x$)?

Absolutely! You can use $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$. If the large sphere moves by $\Delta x$ to the right, the small sphere moves by $10R$ to the left relative to the large sphere, so its absolute displacement is $\Delta x - 10R$. Thus, $4M(\Delta x) + M(\Delta x - 10R) = 0 \implies 5M \Delta x = 10MR \implies \Delta x = 2R$. The new position is $L + 2R$. This is a great shortcut!

Question

A small sphere of radius $R$ is held against the inner surface of a larger sphere of radius $6R$. The masses of large and small spheres are $4M$ and $M$ respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the larger sphere when the smaller sphere reaches the other extreme position.

Sigma Percentile · Accepted Answer

### The Dance of the Spheres: Conservation of Center of Mass

Imagine a massive hollow sphere resting on a perfectly smooth table, and inside it, a smaller sphere is held against its inner wall. When you let the small sphere go, it slides down the curved wall. But here is the fascinating part: as the small sphere moves one way, the large sphere must move the other way. Why? Because the universe demands balance.

### The Hidden Symmetry: Zero External Force

The key to unlocking this problem lies in the word "frictionless." Because there is no friction between the spheres or between the large sphere and the table, there is absolutely **no external force** acting on the system in the horizontal direction.

According to Newton's laws, if the net external force is zero, the acceleration of the center of mass is zero. Since the system starts from rest, the horizontal position of the center of mass ($x_{CM}$) is strictly conserved. It acts as an invisible, immovable anchor point around which the two spheres perform their dance.

### The Initial State: Pinpointing the Center of Mass

Let's map out the initial coordinates. The large sphere (mass $4M$) has its center at $x_1 = L$. The small sphere (mass $M$, radius $R$) is touching the inner surface of the large sphere (radius $6R$) on the right side.

The distance between their centers is the difference in their radii: $6R - R = 5R$. Therefore, the small sphere's center is at $x_2 = L + 5R$.

Using the center of mass formula:
$$x_{CM, i} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

Substituting our values:
$$x_{CM, i} = \frac{(4M)(L) + M(L + 5R)}{4M + M}$$
$$x_{CM, i} = \frac{5ML + 5MR}{5M} = L + R$$

### The Final State: A Shift in Perspective

When the small sphere reaches the "other extreme position," it is now touching the left inner surface of the large sphere. We don't know exactly where the large sphere is now, so let's assign its new center a coordinate $x$.

Because the small sphere is now on the left, its center is $5R$ to the left of the large sphere's center. So, its new coordinate is $x_2' = x - 5R$.

Let's calculate the final center of mass:
$$x_{CM, f} = \frac{(4M)(x) + M(x - 5R)}{5M}$$
$$x_{CM, f} = \frac{5Mx - 5MR}{5M} = x - R$$

### The Grand Equation: Bringing It All Together

Since the center of mass cannot move horizontally, we simply equate the initial and final positions:
$$x_{CM, i} = x_{CM, f}$$
$$L + R = x - R$$

Solving for $x$, we get:
$$x = L + 2R$$

Thus, the final coordinates of the center of the large sphere are **$(L + 2R, 0)$**.

**Pro-Tip (The Displacement Method):** You can also solve this using relative displacements! If the large sphere moves by $\Delta x$, the small sphere moves by $-10R$ relative to the large sphere. Its absolute displacement is $\Delta x - 10R$. Conserving momentum: $4M(\Delta x) + M(\Delta x - 10R) = 0$, which instantly gives $5M\Delta x = 10MR \implies \Delta x = 2R$. Adding this to the initial position $L$ gives $L + 2R$. Physics is beautiful when you see the shortcuts!

Visualized Solution

Initial Setup

Conservation of Centre of Mass

Initial Coordinates

Initial Position of CM

The Final State

Final Coordinates

Final Position of CM

Equating CM Positions

The Sigma Insight: Motion of Centre of Mass

The Dance of the Spheres

The Hidden Symmetry

The Initial State

The Final State

The Grand Equation

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Visualized Solution

Initial Setup

Conservation of Centre of Mass

Initial Coordinates

Initial Position of CM

The Final State

Final Coordinates

Final Position of CM

Equating CM Positions

The Sigma Insight: Motion of Centre of Mass

The Dance of the Spheres

The Hidden Symmetry

The Initial State

The Final State

The Grand Equation

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