- 1] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through a center.
- 2] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through one end.
- C] Moment of Inertia of a circular ring about a perpendicular axis passing through a center
- D] The moment of inertia of a circular disk about an axis perpendicular to its plane and passing through a center.

## 1] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through a center.

Let us consider a thin uniform rod of mass ‘M’ and length ‘L’ rotating about a perpendicular axis and passing through a center. Consider a small element of thickness ‘dx’ of mass ‘dm’ at a distance ‘x’ from the axis of rotation.

Then, Mass of rod per unit length =

mass of small element of thickness ‘‘ is:

As we know the moment of inertia of a particle can be calculated by the product of mass and perpendicular distance from the axis of rotation. So,

Moment of inertia of small element is:

Since the total moment of inertia of a rod can be calculated by adding M.I of all elements, we can do so by integrating with a limit from one end to another end.

**The moment of inertia of a thin rod about the given axis is:**

**If ‘K’ is the radius of gyration**, then we can write:

## 2] Moment of Inertia of the thin rod about an axis perpendicular to its plane and passing through one end.

Let us consider a thin uniform rod of mass ‘M’ and length ‘L’ rotating about a perpendicular axis and passing through one end as shown in the figure above. Consider a small element of thickness ‘dx’ of mass ‘dm’ at a distance ‘x’ from the axis of rotation.

Then, Mass of rod per unit length =

mass of small element of thickness ‘‘ is:

As we know the moment of inertia of a particle can be calculated by the product of mass and perpendicular distance from the axis of rotation. So,

Moment of inertia of small element is:

Since the total moment of inertia of a rod can be calculated by adding M.I of all elements, we can do so by integrating with a limit from one end to another end.

**The moment of inertia of a thin rod about the given axis is:**

**If ‘K’ is the radius of gyration** then,

**Alternative way**

we can use theorem of parallel axis on the moment of inertia to find the moment of inertia of a thin rod about a perpendicular axis passing through at one end. According to the theorem of parallel axis:

where = moment of inertia about an axis through a center of mass (i.e. center of rod) and ‘D’ is the perpendicular distance between the given axis and parallel axis.

Since we know the moment of inertia of a thin rod about perpendicular distance passing through the center is So, and the distance between two parallel axes becomes .

Hence,

## C] Moment of Inertia of a circular ring about a perpendicular axis passing through a center

Consider a thin circular ring of mass ‘M’ and radius ‘R’ rotating around a perpendicular axis and passing through a center as shown in the figure. As shown in the figure, let’s consider a small element of mass ‘dm’.

Moment of inertia of small element

Then, the moment of inertia of the whole circular ring around that axis can be calculated by integrating over the limit.

If ‘K’ is the radius of gyration. then,

The radius of gyration in a circular ring is equal to the radius of a circular ring.

## D] The moment of inertia of a circular disk about an axis perpendicular to its plane and passing through a center.

Consider a thin circular disk of Mass ‘M’ and radius ‘R’ about a perpendicular axis passing through a center. We can consider a circular disk made up of many concentric circular rings. Consider one circular ring of thickness ‘dx’ at a distance ‘x’ from the axis of rotation.

Mass per unit area of a circular disk is

The area of considered ring is

Since is small so, becomes smaller and can be neglected.

then, the mass of circular ring = mass per unit area of a disk

Moment of inertia of a circular ring:

Then, the Moment of inertia of the circular disk about the given axis can be calculated using integrating within the limit.

If ‘K’ is the radius of gyration then,

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