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Step-by-Step Solution
Step 1: Understand the Problem
We have a uniform cylinder of length $l$ and radius $R$, and we want to find the value of the ratio $\frac{l}{R}$ that minimizes its moment of inertia about a perpendicular bisector (an axis through the center, perpendicular to its length).
Step 2: Write the Moment of Inertia Expression
The moment of inertia $I$ of a uniform cylinder about its perpendicular bisector is the sum of two terms:
$I = \frac{M l^2}{12} + \frac{M R^2}{4}$
where $M$ is the mass of the cylinder, $l$ its length, and $R$ its radius.
Step 3: Relate $R$ and $l$ Using the Volume
The volume of the cylinder is
$V = \pi R^2 \, l$.
Because the cylinder is uniform, its volume (and mass) remains constant. Hence:
$R^2 = \frac{V}{\pi \, l}.
Step 4: Substitute $R^2$ into the Moment of Inertia Formula
Substitute $R^2 = \frac{V}{\pi l}$ into $I$:
$I = \frac{M l^2}{12} + \frac{M R^2}{4}
= \frac{M l^2}{12} + \frac{M}{4} \left(\frac{V}{\pi \, l}\right).
So,
$I = \frac{M l^2}{12} + \frac{M V}{4 \pi \, l}.
Note that $M$ and $V$ are constants. Therefore, $I$ is a function of $l$ only.
Step 5: Differentiate to Find the Minimum
To find the extremum (maximum or minimum), we set the derivative of $I$ with respect to $l$ to zero:
$\frac{\mathrm{d}I}{\mathrm{d}l} = 0.
First, compute the derivative:
$\frac{\mathrm{d}I}{\mathrm{d}l}
= \frac{M}{12} \cdot 2l - \frac{M V}{4 \pi \, l^2}
= \frac{M l}{6} - \frac{M V}{4 \pi \, l^2}.
Set this equal to zero:
$\frac{M l}{6} - \frac{M V}{4 \pi \, l^2} = 0.
Step 6: Solve for $l$ in Terms of $R$
Rearrange to isolate $l$-dependent terms:
$\frac{M l}{6} = \frac{M V}{4 \pi \, l^2}.
We can cancel $M$ from both sides. Recall that $V = \pi R^2 l$. Substituting $V$ into the right side gives:
$\frac{l}{6} = \frac{\pi R^2 l}{4 \pi l^2}
\quad \Longrightarrow \quad
\frac{l}{6} = \frac{R^2}{4 l}.
Rearrange to express $l$ and $R$:
$\frac{R^2}{l^2} = \frac{4}{6} = \frac{2}{3}
\quad \Longrightarrow \quad
\left( \frac{l}{R} \right)^2 = \frac{3}{2}
\quad \Longrightarrow \quad
\frac{l}{R} = \sqrt{\frac{3}{2}}.
Step 7: State the Required Ratio
The ratio of $\frac{l}{R}$ that minimizes the moment of inertia is
$\sqrt{\frac{3}{2}}$.
