# How to find the moment of inertia of a solid cylinder about transverse (perpendicular) axis passing through its center?

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The solution is of one line -

# I = (MR^2)/4 + (ML^2)/12#

I didn't understand how #(ML^2)/12# comes into play...

The solution is of one line -

I didn't understand how

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### Answer:

It needs to be done in three steps.

1. Stating Moment of Inertia of a infinitesimally thin Disk.

2. Application of Perpendicular Axis and Parallel axis Theorems.

3. Integrating over the length of the cylinder.

But first of all let's state the problem.

#### Explanation:

**Figure 1**.

Let us consider a cylinder of length

We know that its density

**Figure 2**.

Let us consider that the cylinder is made up of infinitesimally thin disks each of thickness

or

since

or

**Step 1**.

We know that moment of inertia of a circular disk of mass

**Step 2**.

Observe from figure 2, that this moment of inertia has been calculated about

The moment of inertia about an axis which is perpendicular to the plane contained by the remaining two axes is the sum of the moments of inertia about these two perpendicular axes, through the same point in the plane of the object.

It follows that

Also from symmetry we see that moment of inertia about

Combining the equations (3) and (4) we obtain

or

Let the infinitesimal disk be located at a distance

Now we make use of the parallel axis theorem about the

The moment of inertia about any axis parallel to that axis through the center of mass is given by

where

**Step 3**.

Insert the value of

ignoring constant of integration because of it being definite integral.

or

or

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