- Moment of inertia of a system about axis of rotation is given as

$I=\sum m_i r_i^2$

where $m_i$ is the mass of the ith particle and $r_i$ is its perpendicular distance from the axis of rotation

- For a system consisting of collection of discrete particles ,above equation can be used directly for calculating the moment of inertia
- For continuous bodies ,moment of inertia about a given line can be obtained using integration technique
- SI unit of Moment of inertia is $Kg m^2$
- It is a scalar quantity

1. Moment of Inertia calculator for a mass m at distance d from axis of Rotation is given by

$I = md^2$

Where

m -> mass

d -> distance d from axis of Rotation

I -> Moment of inertia

2. Moment of Inertia for a thin rectangular rod around perpendicular bisector and perpendicular axis through one end is given by

$I_p = \frac {mL^2}{12}$

$I_e = \frac {mL^2}{3}$

Where

m -> Mass of the thin rectangular rod

L -> Length of the Rod

$I_p$ -> Moment of Inertia around perpendicular bisector

$I_e$ -> Moment of Inertia around perpendicular axis through one end

3. Moment of Inertia for a solid and hollow sphere about the axis through its center is given by

$I_S = \frac {2}{5}mR^2$

$I_H = \frac {2}{3}mR^2$

Where

m -> Mass of the Sphere

R -> Radius of the sphere

$I_S$ -> Moment of Inertia of Solid sphere about the axis through its center

$I_H$ -> Moment of Inertia of Hollow sphere about the axis through its center

4.Moment of Inertia calculator for a thin or solid cylinder/disk is given

Moment of Inertia for Solid Disk or cylider about the central axis

$I_S = \frac {1}{2}mR^2$

Moment of Inertia for thin Disk or cylider about the central axis

$I_H = mR^2$

- Enter the values of mass and distance

- Click on the calculate button.

$I = md^2$

A particle of mass 5 kg is placed at a distance 5 m from the axis of rotation. Find the moment of inertia of particle about the axis of rotation

m = 5 kg, d=5 m, I=?

From the formula

$I = md^2$

$I=5 \times 5^2 = 125 \ kg m^2$

A object of mass 2 kg is situated at a distance 10 m from the axis of rotation. Find the moment of inertia of about about the axis of rotation

m = 2 kg, d=10 m, I=?

From the formula

$I = md^2$

$I=2 \times 10^2 = 200 \ kg m^2$

- Enter the values of mass and length of the rod

- Click on the calculate button.

$I_p = \frac {mL^2}{12}$

$I_e = \frac {mL^2}{3}$

Calculate the Moment of inertia of thin rectangular rod of Mass 2 kg and Length 2 m around the perpendicular bisector and perpendicular axis through one end

m = 2 kg, L=2 m, $I_e$=?,$I_p$=?

Moment of inertia around perpendicular bisector is given by $I_p = \frac {mL^2}{12}$

$I_p = \frac {2 \times 2^2}{12}= 2/3 \ Kg m^2$br> Moment of inertia about perpendicular axis through one end is given by $I_e = \frac {mL^2}{3}$

$I_e = \frac {2 \times 2^2}{3} = 8/3 \ Kg m^2$br>

A thin rectangular rod has Mass 1 kg and Length 1 m . Find the moment of inertia around the perpendicular bisector and perpendicular axis through one end

m = 1 kg, L=1 m, $I_e$=?,$I_p$=?

Moment of inertia around perpendicular bisector is given by $I_p = \frac {mL^2}{12}$

$I_p = \frac {1 \times 1^2}{12}= 1/12 \ Kg m^2$br> Moment of inertia about perpendicular axis through one end is given by $I_e = \frac {mL^2}{3}$

$I_e = \frac {1 \times 1^2}{3} = 1/3 \ Kg m^2$br>

- Enter the values of mass and radius of the sphere

- Click on the calculate button.

$I_S = \frac {2}{5}mR^2$

$I_H = \frac {2}{3}mR^2$

Calculate the Moment of inertia of hollow sphere of Mass 2 kg and Diameter 6 m about the axis through its center

m = 2 kg, D=6 m, $I$=? Therefore R=3 m Moment of inertia about the axis through its center is given by $I_H = \frac {2}{3}mR^2$

$I_H = \frac {2}{3} \times 2 \times 3^2= 12 \ kg-m^2$

A Solid Sphere has Mass 1 kg and Radius 5 m . Find the moment of inertia about the axis through its center

m = 1 kg,R=5 m, I=?

Moment of inertia about the axis through its center is given by $I_S = \frac {2}{5}mR^2$

$I_S = \frac {2}{5} \times 1 \times 5^2= 10 \ kg-m^2$

- Enter the values of mass and radius of the sphere

- Click on the calculate button.

Moment of Inertia for Solid Disk or cylinder about the central axis

$I_S = \frac {1}{2}mR^2$

Moment of Inertia for thin Disk or cylinder about the central axis

$I_H = mR^2$

Calculate the Moment of inertia of thin cylinder of Mass 10 kg , Diameter .6 m and Length =1 m about the central axis

m = 10 kg, D=.6 m, L=1 m,$I$=? Therefore R=.3 m Moment of Inertia for thin Disk or cylinder about the central axis

$I_H = mR^2$

$I_H = 10 \times .3^2= .9 \ kg-m^2$

A Solid disk has Mass 1 kg and Radius .2 m . Find the moment of inertia about the central axis

m = 1 kg,R=.2 m, I=?

Moment of Inertia for Solid Disk or cylinder about the central axis

$I_S = \frac {1}{2}mR^2$

$I_S = \frac {1}{2} \times 1 \times (.2)^2= .02 \ kg-m^2$

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