Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is M, radius of i

Q: Why do we use slant height $l$ instead of vertical height $y$ for integration?

Integrating along the slant height $l$ directly aligns with the surface area where the mass is actually distributed. If you use $y$, you must remember that the width of the ring is $dy/\cos\theta$, not just $dy$, which often leads to silly mistakes!

Q: Why is the moment of inertia of a hollow cone the same as a solid disc?

It's a beautiful geometric coincidence! If you squash a hollow cone down flat along its axis, all its mass elements maintain their same distance $r$ from the axis. The resulting squashed shape is exactly a solid disc of radius $R$ and mass $M$.

Q: What if the cone was solid instead of hollow?

For a solid cone, the mass is distributed throughout its volume. You would slice it into solid discs instead of rings. The integration would yield $I = \frac{3}{10}MR^2$.

Q: How did we get the area of the elemental ring as $2\pi r dl$?

Imagine cutting the thin ring and unrolling it. It forms a long rectangle. The length of this rectangle is the circumference of the ring, $2\pi r$, and its width is the slant thickness $dl$. So, Area = length $\times$ width = $2\pi r dl$.

Q: Can I solve this using perpendicular axis theorem?

No, the perpendicular axis theorem is strictly for 2D planar laminas. A hollow cone is a 3D object (even though its surface is a 2D shell, it exists in 3D space). You must use integration from first principles.

Question

Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is $M$, radius of its top $R$ and height $H$, then its moment of inertia about its axis is

Options:

(A) $\frac{MR^2}{2}$

(B) $\frac{M(R^2+H^2)}{4}$

(C) $\frac{MH^2}{3}$

(D) $\frac{MR^2}{3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Hollow Cone

Imagine an empty ice cream cone. To find its moment of inertia, we first need to understand its geometry. Let's assume its mass is $M$, base radius is $R$, and height is $H$.

By Pythagoras theorem, its slant height $L$ will be $L = \sqrt{R^2 + H^2}$. If we consider the semi-vertical angle at the vertex as $	heta$, then from the right-angled triangle, $\sin	heta = \frac{R}{L}$.

### Surface Mass Density

Because this cone is hollow inside, all its mass is distributed only over its curved surface. We need to calculate the surface mass density, which we call $\sigma$.

The curved surface area of a cone is $\pi R L$. Therefore, the surface mass density will be the total mass $M$ divided by this area:
$$ \sigma = \frac{M}{\pi R L} $$

### Slicing into Elemental Rings

To find the total moment of inertia, we will slice this entire cone into tiny elemental rings parallel to the base. Suppose at a slant distance $l$ from the vertex, we take a very thin elemental ring of slant thickness $dl$.

From the geometry of similar triangles, the radius $r$ of this ring will be $r = l \sin	heta$. If we unroll this thin ring, it forms a rectangle. Its area $dA$ will be its circumference multiplied by its thickness:
$$ dA = 2\pi r dl = 2\pi (l \sin	heta) dl $$

### Mass of the Elemental Ring

What will be the mass $dm$ of this tiny ring? We simply multiply the surface mass density $\sigma$ with the area $dA$.

$$ dm = \sigma dA = \left(\frac{M}{\pi R L}ight) (2\pi l \sin	heta dl) $$

Now, look closely at the geometry. We know that $R = L \sin	heta$. If we substitute this into our equation, the $\pi$ and $\sin	heta$ terms beautifully cancel out!

$$ dm = \left(\frac{M}{\pi (L \sin	heta) L}ight) (2\pi l \sin	heta dl) = \frac{2M}{L^2} l dl $$

This is a very elegant and simplified expression for the mass of our elemental ring.

### Moment of Inertia of the Ring

The moment of inertia of a ring about its central axis is simply its mass multiplied by the square of its radius. So, for our elemental ring, $dI = dm \cdot r^2$.

Let's substitute the values of $dm$ and $r$:
$$ dI = \left(\frac{2M}{L^2} l dlight) (l \sin	heta)^2 $$

Squaring the radius gives us $l^2 \sin^2	heta$. Multiplying them together, we get:
$$ dI = \frac{2M \sin^2	heta}{L^2} l^3 dl $$

This is the tiny contribution of our elemental ring to the total moment of inertia.

### Integration for the Total Moment of Inertia

To find the moment of inertia of the entire hollow cone, we need to add up the contributions of all such tiny rings. In the language of calculus, we integrate $dI$ from the vertex where $l = 0$, all the way to the base where $l = L$.

$$ I = \int_0^L \frac{2M \sin^2	heta}{L^2} l^3 dl $$

The terms $\frac{2M \sin^2	heta}{L^2}$ are constants, so they come out of the integral. We just need to integrate $l^3 dl$, which is a standard integral giving $\frac{l^4}{4}$.

$$ I = \frac{2M \sin^2	heta}{L^2} \left[ \frac{l^4}{4} ight]_0^L = \frac{2M \sin^2	heta}{L^2} \frac{L^4}{4} $$

Simplifying this, we arrive at:
$$ I = \frac{M L^2 \sin^2	heta}{2} $$

### The Grand Reveal

We are almost there! Our expression is in terms of the slant height $L$ and the angle $	heta$, but the options are in terms of the base radius $R$.

Remember our initial geometric relation: $R = L \sin	heta$. Therefore, $L^2 \sin^2	heta$ is exactly $R^2$! Substituting this back, we get our final, beautiful result:

$$ I = \frac{MR^2}{2} $$

**Notice how this is exactly the same as the moment of inertia of a solid disc!** This is a fascinating geometric coincidence and a highly useful result to remember for your exams.

Animated Solution for Physics - Rotational Motion: Shown in the figure is a hollow icecream cone (it is open at the top). If its mass is $M$ , radius of its top $R$ and height $H$ , then its moment of inertia about its axis is

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Visualized Solution

$Geometry of the Hollow Cone$

$Surface Mass Density$

$Elemental Ring$

$Mass of the Elemental Ring$

$Moment of Inertia of the Ring$

$Integration for Total Moment of Inertia$

$Final Substitution$

The Sigma Insight: Moment of Inertia

The Geometry of the Hollow Cone

Surface Mass Density

Slicing into Elemental Rings

Mass of the Elemental Ring

Moment of Inertia of the Ring

Integration for the Total Moment of Inertia

The Grand Reveal

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the Hollow Cone

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

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

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

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

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

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