Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its c

Q: Why do we need to eliminate $L$ from the moment of inertia equation?

To find the minimum of a function using basic calculus, it's easiest to have it in terms of a single variable. Since the mass is constant, $L$ and $R$ are dependent on each other. Changing one forces the other to change. Eliminating $L$ allows us to see exactly how $I$ changes as we vary $R$ alone.

Q: Why is the mass considered constant?

The problem explicitly states 'If such a cylinder is to be made for a given mass of a material'. This means we have a fixed, finite amount of 'stuff' to build our cylinder, so $M$ cannot change.

Q: What does $\frac{dI}{dR} = 0$ mean physically?

If you were to graph the moment of inertia $I$ against the radius $R$, the curve would dip down to a lowest point and then go back up. At that exact lowest point (the minimum), the slope of the tangent line is perfectly horizontal, which mathematically means the derivative is zero.

Q: Can we differentiate with respect to $L$ instead of $R$?

Absolutely! You would use the mass equation to eliminate $R$ instead of $L$, making $I$ a function of $L$. Differentiating and setting to zero will lead you to the exact same final ratio of $L/R$.

Q: Why did we substitute $M$ back into the equation at the very end?

We reached $R^6 = \frac{2M^2}{3 \rho^2 \pi^2}$. We could have solved for $R$, then plugged it into the $L$ equation, and then divided them. However, substituting $M = \rho \pi R^2 L$ back in is a clever algebraic shortcut that immediately brings $L$ back into the equation and cancels out all the messy constants, leading directly to the ratio.

Question

Moment of inertia of a cylinder of mass $M$, length $L$ and radius $R$ about an axis passing through its centre and perpendicular to the axis of the cylinder is $I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right)$. If such a cylinder is to be made for a given mass of a material. To have minimum possible moment of inertia, the ratio $L/R$ for cylinder is

Options:

(A) $\frac{2}{3}$

(B) $\frac{3}{2}$

(C) $\sqrt{\frac{3}{2}}$

(D) $\sqrt{\frac{2}{3}}$

Sigma Percentile · Accepted Answer

### The Optimization Challenge

Imagine you are tasked with designing a solid cylinder out of a fixed amount of material (a given mass $M$). Your goal is to make this cylinder as easy to rotate as possible about an axis passing through its center and perpendicular to its length. In physics terms, you want to **minimize its moment of inertia**, $I$.

The moment of inertia for this specific axis is given by the formula:
$$I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right)$$

At first glance, it seems like we should just make both $R$ and $L$ as small as possible. But there's a catch! Because the mass is fixed, you can't change $R$ without changing $L$. If you make the cylinder thinner (smaller $R$), it must become longer (larger $L$) to maintain the same mass. We need to find the perfect balance.

### Setting Up the Variables

To solve this optimization problem, we need to express the moment of inertia in terms of a single variable. Let's use the fact that the mass $M$ is constant. The mass of a cylinder is its density $\rho$ multiplied by its volume:
$$M = \rho \pi R^2 L$$

From this, we can express the length $L$ in terms of the radius $R$:
$$L = \frac{M}{\rho \pi R^2}$$

Now, let's substitute this expression for $L$ back into our moment of inertia formula:
$$I = M \left( \frac{R^2}{4} + \frac{1}{12} \left( \frac{M}{\rho \pi R^2} \right)^2 \right)$$
$$I = M \left( \frac{R^2}{4} + \frac{M^2}{12 \rho^2 \pi^2 R^4} \right)$$

Now, $I$ is a function of only one variable, the radius $R$.

### The Calculus of Minimization

To find the minimum value of $I$, we need to take its derivative with respect to $R$ and set it equal to zero. This is where the magic of calculus comes in!

$$\frac{dI}{dR} = \frac{d}{dR} \left[ M \left( \frac{R^2}{4} + \frac{M^2}{12 \rho^2 \pi^2 R^4} \right) \right] = 0$$

Applying the power rule, we get:
$$M \left( \frac{2R}{4} + \frac{M^2}{12 \rho^2 \pi^2} (-4R^{-5}) \right) = 0$$

Since the mass $M$ is not zero, the expression inside the parentheses must be zero. Let's simplify and rearrange the terms:
$$\frac{R}{2} - \frac{4M^2}{12 \rho^2 \pi^2 R^5} = 0$$
$$\frac{R}{2} = \frac{M^2}{3 \rho^2 \pi^2 R^5}$$

Cross-multiplying gives us a clean equation for $R^6$:
$$R^6 = \frac{2M^2}{3 \rho^2 \pi^2}$$

### Unveiling the Optimal Ratio

We have found the condition for minimum moment of inertia, but the question asks for the ratio $L/R$. We could solve for $R$ and then find $L$, but there is a much more elegant algebraic trick. Let's substitute our original mass equation ($M = \rho \pi R^2 L$) back into our result:

$$R^6 = \frac{2 (\rho \pi R^2 L)^2}{3 \rho^2 \pi^2}$$
$$R^6 = \frac{2 \rho^2 \pi^2 R^4 L^2}{3 \rho^2 \pi^2}$$

Notice how beautifully the density $\rho$ and $\pi$ cancel out!
$$R^6 = \frac{2}{3} R^4 L^2$$

Dividing both sides by $R^4$, we get:
$$R^2 = \frac{2}{3} L^2$$

Rearranging to find the ratio of $L^2$ to $R^2$:
$$\frac{L^2}{R^2} = \frac{3}{2}$$

Finally, taking the square root of both sides reveals the optimal ratio:
**$$\frac{L}{R} = \sqrt{\frac{3}{2}}$$**

This tells us that to make the cylinder easiest to spin about its central perpendicular axis, its length should be exactly $\sqrt{1.5}$ (or about 1.22) times its radius.

Select Answer:

Visualized Solution

Visualizing the Setup

The Mass Constraint

Expressing $I$ in a Single Variable

Differentiating for Minimum $I$

Simplifying the Derivative

Re-substituting Mass

Finding the Ratio

The Way Forward

The Sigma Insight: Moment of Inertia

The Optimization Challenge

Setting Up the Variables

The Calculus of Minimization

Unveiling the Optimal Ratio

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