A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the end of light

Q: Why do we use the Parallel Axis Theorem here?

Because the axis of rotation does not pass through the center of mass of the spheres. It passes through the midpoint of the rod, which is parallel to the spheres' own central axes.

Q: What is the moment of inertia of a solid sphere?

For a solid sphere of mass $M$ and radius $R$, the moment of inertia about its central axis is $\frac{2}{5}MR^2$.

Q: Why did we ignore the rod's moment of inertia?

The problem explicitly states it is a "light rod," which in physics terminology means we can assume its mass is negligible (zero) compared to the massive spheres.

Q: Why did we convert 50 cm to 0.5 m?

To maintain consistency in SI units. The mass is in kg, and the final answer is in $\text{kgm}^2$, so all distances must be converted to meters before calculation.

Q: What if the spheres were hollow?

If they were hollow spheres (spherical shells), their moment of inertia about their own center would be $\frac{2}{3}MR^2$ instead of $\frac{2}{5}MR^2$. The rest of the parallel axis theorem application would remain exactly the same.

Question

A system consists of two identical spheres each of mass $1.5 	ext{ kg}$ and radius $50 	ext{ cm}$ at the end of light rod. The distance between the centres of the two spheres is $5 	ext{ m}$. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its mid-point ?

Options:

(A) $18.75 	ext{ kgm}^2$

(B) $1.905 	imes 10^5 	ext{ kgm}^2$

(C) $19.05 	ext{ kgm}^2$

(D) $1.875 	imes 10^5 	ext{ kgm}^2$

Sigma Percentile · Accepted Answer

### Visualizing the System

Imagine a classic dumbbell setup: two identical solid spheres, each of mass $M = 1.5 	ext{ kg}$ and radius $R = 50 	ext{ cm}$, connected by a light rod of length $l = 5 	ext{ m}$. We are tasked with finding the moment of inertia of this entire system about an axis passing right through the middle of the rod, perpendicular to it.

The first crucial detail to notice is the word **"light rod."** In physics, this is a code word meaning the rod's mass is negligible. Therefore, the rod itself contributes nothing to the total moment of inertia. We only need to worry about the two spheres.

### The Power of the Parallel Axis Theorem

To find the total moment of inertia, we must first recall the moment of inertia of a single solid sphere about an axis passing through its own center of mass. This is a standard formula:

$$I_{	ext{cm}} = \frac{2}{5}MR^2$$

However, our system's axis of rotation does not pass through the center of the spheres. It passes through the midpoint of the rod. This means the axis of rotation is parallel to the central axis of each sphere, separated by a distance $d = \frac{l}{2}$.

This is the perfect scenario for the **Parallel Axis Theorem**, which states:

$$I = I_{	ext{cm}} + Md^2$$

Applying this to one of our spheres, its moment of inertia about the central axis becomes:

$$I_{	ext{sphere}} = \frac{2}{5}MR^2 + M\left(\frac{l}{2}ight)^2$$

Since we have two identical spheres symmetrically placed, the total moment of inertia of the system is simply twice this value:

$$I_{	ext{system}} = 2 \left[ \frac{2}{5}MR^2 + \frac{Ml^2}{4} ight]$$

### Crunching the Numbers

Now, let's carefully plug in our values. Before we do, we must ensure all units are consistent. The mass $M$ is $1.5 	ext{ kg}$, and the length $l$ is $5 	ext{ m}$. But the radius $R$ is given as $50 	ext{ cm}$. We must convert this to meters:

$$R = 50 	ext{ cm} = 0.5 	ext{ m}$$

Substituting these into our factored equation:

$$I_{	ext{system}} = 2M \left[ \frac{2}{5}R^2 + \frac{l^2}{4} ight]$$

$$I_{	ext{system}} = 2(1.5) \left[ \frac{2}{5}(0.5)^2 + \frac{5^2}{4} ight]$$

Let's compute the terms inside the bracket step-by-step. First, the rotational inertia term of the sphere itself:

$$\frac{2}{5}(0.5)^2 = 0.4 	imes 0.25 = 0.1$$

Next, the parallel axis shift term:

$$\frac{5^2}{4} = \frac{25}{4} = 6.25$$

### The Final Spin

Adding those up, we get $6.35$ inside the bracket. Multiplying by the outside factor of $3$ (since $2 	imes 1.5 = 3$):

$$I_{	ext{system}} = 3 	imes [0.1 + 6.25]$$

$$I_{	ext{system}} = 3 	imes 6.35 = 19.05 	ext{ kgm}^2$$

Our final moment of inertia comes out to be exactly $19.05 	ext{ kgm}^2$. This perfectly matches option (c). Always remember to check if the connecting elements in such problems have mass; if they do, you simply add their moment of inertia to the final sum!

Select Answer:

Visualized Solution

$I_{system} = ?$

$Parallel Axis Theorem$

$I_{system} = 2 \times I_{sphere}$

$Substitution$

$Calculation$

$Final Answer$

$What if the rod had mass?$

The Sigma Insight: Moment of Inertia

Visualizing the System

The Power of the Parallel Axis Theorem

Crunching the Numbers

The Final Spin

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isystem​=?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Axis Theorem

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isystem​=2×Isphere​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the rod had mass?