A rod of linear mass density $\lambda$ and length $L$ is bent to form a ring of radius $R$. Moment of inertia of ring about any of its diameter is:
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Solution Steps
1
Determine Total Mass of the Rod
Given the linear mass density is $\lambda$ and length is $L$.
Total mass of the rod, $M = \text{Linear Density} \times \text{Length} = \lambda L$.
2
Relate Length to Radius
The rod is bent into a ring of radius $R$. Thus, the circumference of the ring equals the length of the rod:
$$L = 2\pi R \implies R = \frac{L}{2\pi}$$
3
Moment of Inertia about Central Axis
The moment of inertia ($I_z$) of a ring about an axis passing through its center and perpendicular to its plane is:
$$I_z = MR^2$$
4
Apply Perpendicular Axis Theorem
For a planar object like a ring, let $I_d$ be the moment of inertia about a diameter. By the perpendicular axis theorem:
$$I_z = I_x + I_y = I_d + I_d = 2I_d$$
$$\therefore I_d = \frac{1}{2}I_z = \frac{1}{2}MR^2$$
5
Substitute $M$ and $R$
Substitute $M = \lambda L$ and $R = \frac{L}{2\pi}$ into the equation for $I_d$:
$$I_d = \frac{1}{2}(\lambda L)\left(\frac{L}{2\pi}\right)^2 = \frac{1}{2}(\lambda L)\left(\frac{L^2}{4\pi^2}\right)$$
$$I_d = \frac{\lambda L^3}{8\pi^2}$$
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Theory
1. Linear Mass Density ($\lambda$)
Linear mass density represents how mass is distributed along a one-dimensional object like a wire or a thin rod. It is mathematically expressed as $\lambda = \frac{dm}{dl}$. For a uniform rod, the total mass $M$ is simply the product of its length $L$ and $\lambda$. This concept is fundamental in rotational mechanics because it allows us to express inertia terms in terms of geometric dimensions and material properties. In this problem, it bridges the gap between the physical rod and the mathematical ring.
2. Moment of Inertia of a Ring
The Moment of Inertia (MOI) measures an object’s resistance to rotational acceleration about a specific axis. For a thin ring of mass $M$ and radius $R$, if the axis passes through the center perpendicular to the plane, all mass elements are at the same distance $R$ from the axis, resulting in $I = \int R^2 dm = MR^2$. Understanding how mass distribution relative to the axis affects $I$ is key to solving rigid body dynamics problems.
3. Perpendicular Axis Theorem
This theorem states that for any planar body, the moment of inertia about an axis perpendicular to the plane ($I_z$) is the sum of the moments of inertia about two mutually perpendicular axes lying in the plane ($I_x$ and $I_y$) that intersect at the same point: $I_z = I_x + I_y$. For a symmetric object like a ring, any diameter is equivalent, so $I_x = I_y = I_d$, which leads to the result that the diameter’s MOI is exactly half of the central perpendicular MOI.
4. Geometric Transformations in Mechanics
Many competitive physics problems involve “reforming” objects, such as bending a rod into a circle or a square. The core constraint in these transformations is the conservation of mass and length. Here, the rod’s length $L$ becomes the ring’s circumference $2\pi R$. By establishing this geometric link, we can express the final rotational property (MOI) using the initial parameters ($\lambda, L$) provided in the problem statement, a common technique in JEE Main Physics.
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FAQs
1
What if the rod was bent into a square?
The side length would be $L/4$, and you would calculate the MOI of the four rods about the center using the parallel axis theorem.
2
Does the thickness of the rod matter?
For a “thin” rod, we assume thickness is negligible compared to length and radius, so it doesn’t affect the calculation.
3
Why is $I_z = MR^2$ and not $1/2 MR^2$?
$MR^2$ is for a ring where all mass is at distance R. $1/2 MR^2$ is for a solid disk.
4
Can we use the parallel axis theorem here?
Parallel axis theorem is used to find MOI about an axis parallel to one passing through the center of mass. Here, the perpendicular axis theorem is more direct for diameters.
5
What is the MOI about a tangent in the plane?
Using the parallel axis theorem: $I_{tan} = I_d + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
6
What is the MOI about a tangent perpendicular to the plane?
$I_{tan \perp} = I_z + MR^2 = MR^2 + MR^2 = 2MR^2$.
7
How does $\lambda$ relate to volume density $\rho$?
$\lambda = \rho \times A$, where $A$ is the cross-sectional area of the rod.
8
What if the rod was non-uniform?
Then $\lambda$ would be a function of $x$, and you would need to integrate to find the mass and center of mass.
9
Why is the answer in $L^3$?
Dimensionally, MOI is [Mass][Length]². Since $M = \lambda L$, $I \propto (\lambda L)(L)^2 = \lambda L^3$.
10
Is the MOI different for different diameters?
No, due to the circular symmetry of the ring, the MOI about any diameter passing through the center is identical.
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