The radius of gyration of a uniform rod of length l, about an axis passing through a point l/4 away

Q: What exactly is the radius of gyration?

Think of it as a "summary" of how mass is distributed. It's the radial distance from the axis of rotation where you could compress the entire mass of the object into a single point mass, and it would still have the exact same moment of inertia.

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

The Parallel Axis Theorem is used whenever we need to find the moment of inertia about an axis that is parallel to an axis passing through the center of mass. Since our new axis is parallel to the CM axis and shifted by $l/4$, it's the perfect tool.

Q: Can we use the Perpendicular Axis Theorem instead?

No. The Perpendicular Axis Theorem is strictly for 2D planar laminas (like a thin disc or a square plate). A rod is treated as a 1D object, and we are shifting the axis parallel to itself, not finding an axis perpendicular to two others.

Q: Why did the mass $m$ cancel out in the final step?

The radius of gyration is a purely geometric property for a uniform body. It describes the *distribution* of mass relative to the axis, not the total amount of mass. That's why $m$ cancels out!

Q: What would the radius of gyration be if the axis was at the end of the rod?

If the axis is at the end, the distance $d$ is $l/2$. The moment of inertia becomes $I = \frac{ml^2}{12} + m(l/2)^2 = \frac{ml^2}{3}$. Equating to $mk^2$, the radius of gyration would be $k = \frac{l}{\sqrt{3}}$.

Question

The radius of gyration of a uniform rod of length $l$, about an axis passing through a point $\frac{l}{4}$ away from the centre of the rod and perpendicular to it, is

Options:

(A) $\frac{1}{8}l$

(B) $\sqrt{\frac{3}{8}}l$

(C) $\sqrt{\frac{7}{48}}l$

(D) $\frac{1}{4}l$

Sigma Percentile · Accepted Answer

### The Dance of Inertia: Finding the Radius of Gyration

Imagine you are holding a long, heavy pole. If you try to spin it by holding it exactly in the middle, it feels relatively easy. But if you shift your grip towards one end, spinning it suddenly requires much more effort. This resistance to rotational motion is what we call the **Moment of Inertia**.

In this problem, we are exploring a fascinating property related to this rotational inertia known as the **Radius of Gyration**. We have a uniform rod of length $l$, and we want to find its radius of gyration about an axis that is shifted away from its center by a distance of $l/4$.

### The Master Tool: Parallel Axis Theorem

To find the radius of gyration, our first stepping stone is to calculate the actual moment of inertia about this new, shifted axis.

We already know the standard formula for the moment of inertia of a uniform rod about an axis passing through its center of mass (CM):
$$I_{CM} = \frac{ml^2}{12}$$

But our axis isn't at the center; it's shifted parallel to the CM axis by a distance $d = l/4$. This is where the **Parallel Axis Theorem** comes to our rescue. It states that the moment of inertia about any parallel axis is the sum of the moment of inertia about the CM and the product of the mass and the square of the distance between the axes:
$$I = I_{CM} + md^2$$

### Executing the Math

Let's carefully substitute our known values into the theorem. We plug in $d = l/4$:
$$I = \frac{ml^2}{12} + m\left(\frac{l}{4}\right)^2$$

Squaring the fraction gives us $l^2/16$. Now, we just need to add the two terms:
$$I = \frac{ml^2}{12} + \frac{ml^2}{16}$$

To add these fractions, we find the lowest common multiple of 12 and 16, which is 48. Adjusting the numerators, we get:
$$I = \frac{4ml^2 + 3ml^2}{48} = \frac{7ml^2}{48}$$

### Unveiling the Radius of Gyration

Now that we have the moment of inertia, what exactly is the radius of gyration, $k$?

Think of $k$ as a geometric summary of the mass distribution. It is the theoretical distance from the axis where you could compress the entire mass of the rod into a single point, and it would still possess the exact same moment of inertia. Mathematically, it is defined as:
$$I = mk^2$$

We simply equate this definition to the moment of inertia we just calculated:
$$mk^2 = \frac{7ml^2}{48}$$

Notice how the mass $m$ beautifully cancels out from both sides! This proves that the radius of gyration is a purely geometric property. It doesn't matter if the rod is made of lightweight plastic or heavy lead; $k$ will be the same.
$$k^2 = \frac{7l^2}{48}$$

Taking the square root of both sides, we arrive at our final, elegant answer:
$$k = \sqrt{\frac{7}{48}}l$$

This perfectly matches option (c). The next time you spin an object, remember that its resistance to spinning is dictated not just by its mass, but by this hidden geometric distance—the radius of gyration!

Animated Solution for Physics - Rotational Motion: The radius of gyration of a uniform rod of length $l$ , about an axis passing through a point $\frac{l}{4}$ away from the centre of the rod and perpendicular to it, is

Select Answer:

Visualized Solution

Visualizing the Setup

The Parallel Axis Theorem

Substituting the Values

Calculating Moment of Inertia

Defining Radius of Gyration

Solving for $k$

Physical Significance

The Sigma Insight: Moment of Inertia

The Dance of Inertia

The Master Tool

Executing the Math

Unveiling the Radius of Gyration

Similar Questions

A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $(\frac{σ _{0}}{r})$ , then the radius of gyration of the disc about its axis passing through the centre is

The linear mass density of a thin rod $A B$ of length $L$ varies from $A$ to $B$ as $λ (x) = λ_{0} (1 + \frac{x}{L})$ , where $x$ is the distance from $A$ . If $M$ is mass of the rod, then its moment of inertia about an axis passing through $A$ and perpendicular to the rod is

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$ . What is the ratio $l / R$ such that the moment of inertia is minimum?

Moment of inertia of a circular wire of mass $M$ and radius $R$ about its diameter is

One quarter section is cut from a uniform circular disc of radius $R$ . This section has a mass $M$ . It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

A thin wire of length $L$ and uniform linear mass density $ρ$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $X X^{'}$ is

Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2 R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

Moment of inertia of a square plate of side $l$ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by

Match List I with List II.

List-I

List-II

Let the moment of inertia of a hollow cylinder of length $30 cm$ (inner radius $10 cm$ and outer radius $20 cm$ ) about its axis be $I$ . The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also $I$ , is