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?

• A) 3/√2
• B) √(3/2)
• C) √3/2
• D) 1

To find the ratio \( L/R \) that minimizes the moment of inertia \( I \) of a uniform cylinder about its perpendicular bisector, we need to start by recalling the formula for the moment of inertia of a cylinder. The moment of inertia \( I \) about the perpendicular bisector (which is an axis through the center and perpendicular to the length) can be expressed as:

Understanding the Moment of Inertia

The moment of inertia \( I \) for a uniform cylinder can be calculated using the formula:

I = \frac{1}{12} m (3R^2 + L^2)

Here, \( m \) is the mass of the cylinder, \( R \) is the radius, and \( L \) is the length. The mass \( m \) can be expressed in terms of the volume and density:

m = \rho \cdot V = \rho \cdot \pi R^2 L

Substituting this into the moment of inertia formula gives:

I = \frac{1}{12} (\rho \cdot \pi R^2 L) (3R^2 + L^2)

Finding the Ratio L/R

To minimize \( I \), we need to express it in terms of the ratio \( x = \frac{L}{R} \). Thus, we can rewrite \( L \) as \( L = xR \). Substituting this into the moment of inertia formula yields:

I = \frac{1}{12} (\rho \cdot \pi R^2 (xR)) (3R^2 + (xR)^2)

Simplifying this gives:

I = \frac{1}{12} \rho \pi R^4 x (3 + x^2)

Now, we can focus on minimizing the expression \( x(3 + x^2) \). To find the minimum, we can take the derivative with respect to \( x \) and set it to zero:

\frac{d}{dx} [x(3 + x^2)] = 3 + 2x^2 = 0

Solving for \( x \) gives:

2x^2 + 3 = 0

This equation does not yield real solutions, so we need to analyze the function \( f(x) = x(3 + x^2) \) directly. We can find the critical points by examining the second derivative or by testing values.

Evaluating Critical Points

To find the minimum, we can evaluate the function at specific points. Testing \( x = 1 \) and \( x = \sqrt{3/2} \) gives us:

• For \( x = 1 \): \( f(1) = 1(3 + 1) = 4 \)
• For \( x = \sqrt{3/2} \): \( f(\sqrt{3/2}) = \sqrt{3/2}(3 + 3/2) = \sqrt{3/2} \cdot \frac{9}{2} = \frac{9\sqrt{3}}{4} \approx 3.897 \)

From this evaluation, we see that \( x = \sqrt{3/2} \) yields a lower moment of inertia than \( x = 1 \). Thus, the ratio \( L/R \) that minimizes the moment of inertia is:

Final Answer

The correct answer is B) \( \sqrt{3/2} \).