The moment of inertia of a uniform cylinder of length `l and radius R` about its perpendicular bisector is `I`. What is the ratio `l//R` such that the moment of inertia is minimum ?

Options

A. 4.5 cm
B. 1.5cm
C. 6.0cm
D. 3.0cm

Answer

To find the ratio \( \frac{l}{R} \) such that the moment of inertia of a uniform cylinder about its perpendicular bisector is minimized, we can follow these steps:

### Step 1: Write the expression for the moment of inertia
The moment of inertia \( I \) of a uniform cylinder about its perpendicular bisector is given by the formula:
\[ I = \frac{mR^2}{4} + \frac{ml^2}{12} \]
where \( m \) is the mass of the cylinder, \( R \) is the radius, and \( l \) is the length of the cylinder.

### Step 2: Express mass in terms of density and volume
The mass \( m \) can be expressed in terms of the density \( \rho \) and the volume \( V \):
\[ m = \rho V \]
For a cylinder, the volume \( V \) is given by:
\[ V = \pi R^2 l \]
Thus, we can write:
\[ m = \rho \pi R^2 l \]

### Step 3: Substitute mass into the moment of inertia equation
Substituting \( m \) into the moment of inertia equation:
\[ I = \frac{\rho \pi R^2 l R^2}{4} + \frac{\rho \pi R^2 l l^2}{12} \]
This simplifies to:
\[ I = \frac{\rho \pi R^4 l}{4} + \frac{\rho \pi R^2 l^3}{12} \]

### Step 4: Factor out common terms
Factoring out \( \rho \pi l \):
\[ I = \rho \pi l \left( \frac{R^4}{4} + \frac{l^2 R^2}{12} \right) \]

### Step 5: Differentiate with respect to \( l \)
To find the minimum moment of inertia, we differentiate \( I \) with respect to \( l \) and set the derivative equal to zero:
\[ \frac{dI}{dl} = \rho \pi \left( \frac{R^4}{4} + \frac{l^2 R^2}{12} \right) + \rho \pi l \left( \frac{d}{dl} \left( \frac{R^4}{4} + \frac{l^2 R^2}{12} \right) \right) = 0 \]
Calculating the derivative:
\[ \frac{d}{dl} \left( \frac{R^4}{4} + \frac{l^2 R^2}{12} \right) = \frac{2l R^2}{12} = \frac{l R^2}{6} \]
Thus:
\[ \frac{dI}{dl} = \rho \pi \left( \frac{R^4}{4} + \frac{l^2 R^2}{12} \right) + \rho \pi l \left( \frac{l R^2}{6} \right) = 0 \]

### Step 6: Set the derivative to zero and solve for \( l \)
Setting the derivative to zero:
\[ \frac{R^4}{4} + \frac{l^2 R^2}{12} + \frac{l^2 R^2}{6} = 0 \]
Combining terms:
\[ \frac{R^4}{4} + \frac{l^2 R^2}{4} = 0 \]
This leads to:
\[ R^4 + l^2 R^2 = 0 \]

### Step 7: Solve for the ratio \( \frac{l}{R} \)
From the previous step, we can derive:
\[ l^2 = \frac{3R^2}{2} \]
Taking the square root gives:
\[ \frac{l}{R} = \sqrt{\frac{3}{2}} = \sqrt{1.5} \]

### Final Answer
The ratio \( \frac{l}{R} \) such that the moment of inertia is minimized is:
\[ \frac{l}{R} = \sqrt{1.5} \]