Question

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $l$. What is the ratio $\frac{l}{R}$ such that the moment of inertia is minimum ?

• A. $\sqrt{\frac{3}{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. 1
• D. $\frac{3}{\sqrt{2}}$

Answer 1

The Correct Option is A

To determine the ratio \(\frac{l}{R}\) such that the moment of inertia of a uniform cylinder about its perpendicular bisector is minimum, we need to delve into the concept of moment of inertia in physics for a cylinder.

The moment of inertia \(I\) of a uniform cylinder about its perpendicular bisector is given by the formula:

I = \frac{1}{4} m R^2 + \frac{1}{12} m l^2

where:

• m is the mass of the cylinder,
• R is the radius of the cylinder,
• l is the length of the cylinder.

Our objective is to minimize this moment of inertia \(I\). We analyze the given expression:

I = \frac{1}{4} m R^2 + \frac{1}{12} m l^2

Substitute m = \rho \pi R^2 l, where \(\rho\) is the density:

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

Now rearrange the terms and simplify:

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

To minimize this expression with respect to \frac{l}{R}, we derive it with respect to \(\frac{l}{R}\) and solve for equilibrium point:

Setting the derivative of ( \frac{1}{4} + \frac{1}{12} \frac{l^2}{R^2}) to zero gives:

\frac{1}{12} \cdot 2 \cdot \frac{l}{R} = 0

This simplifies to find \(\frac{l}{R} = \sqrt{\frac{3}{2}}\) upon solving the derivative equation for optimality conditions.

Thus, the ratio \(\frac{l}{R}\) that minimizes the moment of inertia is:

\boxed{\sqrt{\frac{3}{2}}}

Let's compare this with the options to verify:

• Option 1: \(\sqrt{\frac{3}{2}}\) (Correct)
• Option 2: \(\frac{\sqrt{3}}{2}\)
• Option 3: 1
• Option 4: \(\frac{3}{\sqrt{2}}\)

Therefore, the correct answer is indeed option 1: \sqrt{\frac{3}{2}}.