Animated Solution for Physics - Rotational Motion: 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?

Visualize this... Imagine a uniform solid cylinder of length $l$ and radius $R$. We are interested in its moment of inertia about an axis passing right through its center, perpendicular to its length.

Look closely at this. The formula for the moment of inertia of a solid cylinder about this perpendicular bisector is given by:

Here, the mass $m$ is constant, but we can change its shape by varying $l$ and $R$. Think of a piece of clay. You can roll it into a long, thin snake or squash it into a flat pancake. Both have the same mass, but their moments of inertia are vastly different.

There is a catch here. To find the minimum moment of inertia, we need to express it in terms of a single variable. Let's use the fact that the mass is constant.

Mass is density times volume, so $m=\rho \pi R^{2}l$. From this, we can write $R^2$ as:

So, let's move forward. Now, let's substitute this expression for $R^2$ back into our moment of inertia formula. This gives us $I$ as a function of $l$ alone:

Is this much clear? We have successfully eliminated $R$ from the equation.

Don't get intimidated. To find the minimum value of $I$, we need to take its derivative with respect to $l$ and set it to zero.

Using the power rule, the derivative of $\frac{1}{l}$ is $-\frac{1}{l^2}$, and the derivative of $l^2$ is $2l$. Setting this to zero gives us our condition for minimum inertia:

By moving the negative term to the other side, we get:

This is the critical condition that must be satisfied. Are you getting the point?

Now look at the equation. To find the ratio of $l$ to $R$, let's substitute the original expression for mass, $m=\rho \pi R^{2}l$, back into our condition. This will help us bring $R$ back into the picture.

It is very simple. Canceling out $\rho$, $\pi$, and one power of $l$ on the left side, we are left with:

Rearranging this gives:

Taking the square root, we get our final answer:

Did you get the feel of it? This ratio gives the optimal shape that minimizes rotational resistance about that specific axis.