Animated Solution for Physics - Rotational Motion: The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of $x$ from it, is '$I(x)$'. Which one of the graphs represents the variation of $I(x)$ with $x$ correctly?

Imagine a solid sphere resting in space. The axis passing directly through its center of mass (COM) is the axis about which the sphere is easiest to rotate. The moment of inertia about this central axis is a constant value, which we denote as $I_C$. For a solid sphere, this value is exactly $\frac{2}{5}M{R}^2$, but for the sake of this graphical analysis, we just need to know that it is a positive constant.

Now, suppose we want to rotate this sphere about a different axis. This new axis is perfectly parallel to the original central axis but is shifted away by a perpendicular distance $x$. How does the resistance to rotation—the moment of inertia—change as we move this axis further and further away?

To find the moment of inertia about this new, shifted axis, we rely on one of the most fundamental and powerful tools in rotational mechanics: The Parallel Axis Theorem.

The theorem states that the moment of inertia $I$ about any axis parallel to an axis through the center of mass is given by:

Here, $M$ is the total mass of the sphere, and $x$ is the perpendicular distance between the two parallel axes. This elegant equation tells us that the moment of inertia is always minimum when the axis passes through the center of mass, and it increases as you move the axis away.

Let's look at the equation $I(x)=I_C+M{x}^2$ through a mathematical lens. If we plot $I(x)$ on the y-axis and $x$ on the x-axis, this equation takes the familiar form of a quadratic function:

Where $c=I_C$ (the y-intercept) and $k=M$ (a positive constant determining the curvature).

Because the highest power of $x$ is $2$, the graph must be a parabola. Furthermore, because the mass $M$ is always a positive quantity, the parabola must open upwards.

Let's establish the boundary conditions to finalize our graph:

1. At $x=0$: The axis is exactly at the center of mass. The equation gives $I(0)=I_C+M(0{)}^2=I_C$. This means the graph does not start at the origin $(0,0)$. Instead, it starts at a positive value $I_C$ on the y-axis. 2. As $x$ increases: The term $M{x}^2$ grows quadratically. The curve sweeps upwards, getting steeper as $x$ gets larger.

When we compare our deduced shape with the given options, we can easily eliminate the straight lines (options a and c) because our relationship is quadratic, not linear. We can also eliminate any curve that starts at the origin or opens downwards. The only graph that perfectly matches an upward-opening parabola starting from a positive y-intercept is Option (b).