Four solid spheres each of diameter √5 cm and mass 0.5 kg are placed with their centres at the corne

Q: Why is the perpendicular distance for spheres 1 and 3 exactly $a/\sqrt{2}$?

The diagonal of a square of side $a$ has a length of $a\sqrt{2}$. The diagonals of a square bisect each other at right angles. Therefore, the distance from any corner to the opposite diagonal is exactly half the length of the diagonal, which is $\frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.

Q: Why don't we use the Parallel Axis Theorem for spheres 2 and 4?

The Parallel Axis Theorem is used when the axis of rotation is at a distance from the center of mass. For spheres 2 and 4, the diagonal axis passes exactly through their centers of mass. Thus, the distance $d$ is zero, and their moment of inertia is simply $I_{cm}$.

Q: I got a different answer because I used the diameter instead of the radius. Why?

The formula for the moment of inertia of a solid sphere is $\frac{2}{5}mr^2$, where $r$ is the radius. The problem provides the diameter as $\sqrt{5}$ cm. You must divide this by 2 to get the radius $r = \frac{\sqrt{5}}{2}$ cm before squaring it.

Q: Can I calculate the moment of inertia about the center of the square first and then shift it?

While you can find the moment of inertia tensor for the system at the center of mass, it's much simpler and less error-prone to calculate the scalar moment of inertia directly about the specified diagonal axis by summing the individual contributions of each sphere.

Q: Why do we treat the spheres as solid spheres and not point masses?

The problem explicitly states they are 'solid spheres' and provides their diameter. If they were point masses, their moment of inertia about their own center of mass would be zero. Because they have a physical size, they possess intrinsic rotational inertia ($\frac{2}{5}mr^2$) that must be accounted for.

Question

Four solid spheres each of diameter $\sqrt{5}$ cm and mass $0.5$ kg are placed with their centres at the corners of a square of side $4$ cm. The moment of inertia of the system about the diagonal of the square is $N 	imes 10^{-4}$ kg-m$^2$, then $N$ is

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine a perfect square resting on a flat plane. At each of its four corners, we place a solid sphere.

These aren't just point masses; they are physical, solid spheres with a defined radius and mass.

The problem asks us to find the moment of inertia of this entire four-sphere system about a very specific axis: the diagonal of the square.

To conquer this, we must look at the geometry of the setup and realize that not all spheres relate to this axis in the same way.

### The Spheres on the Axis

Let's draw a line through one of the diagonals.

Notice what happens to the spheres located at the corners that this diagonal connects. The axis of rotation passes exactly through their centers!

Because the axis passes through their center of mass, we don't need to worry about any offset distance.

The moment of inertia for each of these two spheres is simply the standard formula for a solid sphere rotating about its own diameter.

$$I_{axis} = \frac{2}{5}mr^2$$

Since there are two such spheres, their combined contribution to the total moment of inertia is simply twice this value.

### The Spheres off the Axis: Parallel Axis Theorem

Now, look at the other two spheres. They sit at the remaining two corners of the square, far away from our diagonal axis.

When the system rotates, these spheres will sweep out large circles around the diagonal.

To find their moment of inertia, we must invoke the **Parallel Axis Theorem**. This theorem states that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the center of mass, plus the mass times the square of the perpendicular distance between the axes.

$$I = I_{cm} + md^2$$

But what is this perpendicular distance, $d$?

Geometrically, the distance from a corner of a square to its diagonal is exactly half the length of the diagonal.

Since the side of the square is $a$, the full diagonal is $a\sqrt{2}$. Therefore, the perpendicular distance $d$ is:

$$d = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$$

Plugging this into the Parallel Axis Theorem, the moment of inertia for one of these off-axis spheres is:

$$I_{off-axis} = \frac{2}{5}mr^2 + m\left(\frac{a}{\sqrt{2}}ight)^2 = \frac{2}{5}mr^2 + \frac{ma^2}{2}$$

### The Master Equation

We now have the individual contributions of all four spheres.

The total moment of inertia of the system is simply the algebraic sum of these four individual moments of inertia.

$$I_{total} = 2 	imes \left(\frac{2}{5}mr^2ight) + 2 	imes \left(\frac{2}{5}mr^2 + \frac{ma^2}{2}ight)$$

Let's expand and simplify this expression.

$$I_{total} = \frac{4}{5}mr^2 + \frac{4}{5}mr^2 + ma^2$$

$$I_{total} = \frac{8}{5}mr^2 + ma^2$$

This is our master equation. It beautifully captures the physics of the system before we even touch the numerical values.

### Navigating the Units and Final Calculation

This is where many students stumble. The problem gives us the diameter of the spheres in centimeters and the side of the square in centimeters.

We must convert these to standard SI units (meters) to match the final answer format of $	ext{kg-m}^2$.

The mass is $m = 0.5$ kg.

The diameter is $\sqrt{5}$ cm, so the radius is $r = \frac{\sqrt{5}}{2}$ cm. Converting to meters, we get $r = \frac{\sqrt{5}}{2} 	imes 10^{-2}$ m.

Squaring this gives:

$$r^2 = \frac{5}{4} 	imes 10^{-4} 	ext{ m}^2$$

The side of the square is $a = 4$ cm, which is $4 	imes 10^{-2}$ m.

Squaring this gives:

$$a^2 = 16 	imes 10^{-4} 	ext{ m}^2$$

Now, we carefully substitute these values into our master equation:

$$I_{total} = 0.5 \left[ \frac{8}{5} \left(\frac{5}{4} 	imes 10^{-4}ight) + 16 	imes 10^{-4} ight]$$

Watch how elegantly the numbers cancel out. The $5$ in the numerator and denominator cancel, and $\frac{8}{4}$ simplifies to $2$.

$$I_{total} = 0.5 \left[ 2 	imes 10^{-4} + 16 	imes 10^{-4} ight]$$

$$I_{total} = 0.5 \left[ 18 	imes 10^{-4} ight]$$

$$I_{total} = 9 	imes 10^{-4} 	ext{ kg-m}^2$$

The problem states that the moment of inertia is $N 	imes 10^{-4} 	ext{ kg-m}^2$.

By comparing our result, we can confidently conclude that **$N = 9$**.

Enter Numerical Value:

Visualized Solution

System Setup

Analyzing the Spheres

Moment of Inertia for Spheres on Axis

Distance of Off-Axis Spheres

Parallel Axis Theorem

Total Moment of Inertia

Substituting Values

Final Calculation

Conclusion

The Sigma Insight: Moment of Inertia

Visualizing the Setup

The Spheres on the Axis

The Spheres off the Axis

The Master Equation

Navigating the Units and Final Calculation

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