The moment of inertia of uniform semi-circular disc of mass M and radius r about a line perpendicula

Q: Why is the formula exactly the same as a full disc?

Because the formula $\frac{1}{2}Mr^2$ depends on the mass $M$ of the object itself. The mass in the semi-circle is distributed at the exact same radial distances from the axis as it would be in a full disc. The geometry relative to the axis hasn't changed, only the total amount of mass has.

Q: Does the shift in the center of mass affect this calculation?

No! The axis we are calculating for passes through the geometric center $O$, not the center of mass of the semi-circle. If the question asked for the moment of inertia about an axis passing through the center of mass, we would then have to use the parallel axis theorem.

Q: What if the axis was along the straight diameter edge instead of perpendicular to the plane?

By the perpendicular axis theorem on a full disc, $I_x = I_y = \frac{1}{4}M_{total}r^2$. Applying the same symmetry logic, for the half disc, it would be $\frac{1}{4}Mr^2$.

Q: Can I use integration to solve this instead of the symmetry trick?

Absolutely. Integrating $dI = r^2 dm$ over the semi-circular area yields the exact same result. However, the symmetry trick is a conceptual shortcut that saves precious time in competitive exams.

Q: Does this logic apply to a quarter disc as well?

Yes! A quarter disc of mass $M$ rotating about the center also has a moment of inertia of $\frac{1}{2}Mr^2$. You can imagine four identical quarter discs making up a full disc of mass $4M$. The math works out exactly the same way.

Question

The moment of inertia of uniform semi-circular disc of mass $M$ and radius $r$ about a line perpendicular to the plane of the disc through the centre is

Options:

(A) $\frac{1}{4}Mr^2$

(B) $\frac{2}{5}Mr^2$

(C) $Mr^2$

(D) $\frac{1}{2}Mr^2$

Sigma Percentile · Accepted Answer

### The Intuitive Trap

When faced with a semi-circular disc, the immediate instinct of many students is to dive straight into calculus. They set up an elemental ring, write down $dI = r^2 dm$, and begin integrating over the limits $0$ to $\pi$. While mathematically rigorous, this approach consumes precious time in competitive exams.

There is a far more elegant and powerful way to solve this: **The Principle of Superposition and Symmetry**.

### The Superposition Hack

Imagine you have a semi-circular disc of mass $M$ and radius $r$. The axis of rotation passes right through its geometric center $O$ and is perpendicular to its plane.

Now, let's perform a thought experiment. What if we take another identical semi-circular disc of mass $M$ and attach it right below our original disc?

Suddenly, we have constructed a complete, uniform circular disc!

What is the total mass of this new complete disc? Since it is made of two identical halves, its total mass is simply:

$$M_{total} = M + M = 2M$$

### The Mathematical Proof

We already know the standard formula for the moment of inertia of a complete uniform disc of mass $m$ and radius $r$ about its central axis:

$$I = \frac{1}{2}mr^2$$

Let's substitute the mass of our newly formed complete disc ($m = 2M$) into this standard formula:

$$I_{total} = \frac{1}{2}(2M)r^2 = Mr^2$$

Because moment of inertia is a scalar quantity that simply adds up, the total moment of inertia must be the sum of the moments of inertia of the two halves. Due to perfect symmetry, both halves contribute equally to the total moment of inertia. Let the moment of inertia of one half be $I_1$.

$$I_{total} = I_1 + I_1 = 2I_1$$

Equating the two expressions for $I_{total}$:

$$2I_1 = Mr^2$$

$$I_1 = \frac{1}{2}Mr^2$$

And there we have it! The moment of inertia of the semi-circular disc is exactly $\frac{1}{2}Mr^2$.

**The Golden Takeaway:** This symmetry argument is incredibly versatile. It proves that for *any* sector of a uniform disc (be it a semi-circle, a quarter-circle, or a $60^\circ$ slice) of mass $M$ and radius $r$, the moment of inertia about the central perpendicular axis is always $\frac{1}{2}Mr^2$. The fraction of the circle doesn't change the structure of the formula; it only dictates the mass $M$!

Animated Solution for Physics - Rotational Motion: The moment of inertia of uniform semi-circular disc of mass $M$ and radius $r$ about a line perpendicular to the plane of the disc through the centre is

Select Answer:

Visualized Solution

$Visualizing the Semi-Circular Disc$

$The Symmetry Trick$

$Mass of the Complete Disc$

$Moment of Inertia of Complete Disc$

$Superposition Principle$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Moment of Inertia

The Intuitive Trap

The Superposition Hack

The Mathematical Proof

Similar Questions

A thin disc of mass $M$ and radius $R$ has mass per unit area $σ (r) = k r^{2}$ , where $r$ is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane 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

From a uniform circular disc of radius $R$ and mass $9 M$ , a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is

Mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is

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

Consider a uniform wire of mass $M$ and length $L$ . It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre 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

Consider a uniform square plate of side $a$ and mass $m$ . The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Semi-Circular Disc

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of the Complete Disc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia of Complete Disc

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

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

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