# Why do we use moment of inertia instead of moment of mass?

I am learning Newtonian mechanics in high school. I understand that in rotational motion, the distance between center of mass and the rotational axis has also a role to play. So we find the "moment" of these quantities by multiplying them with $$r$$.

In translational motion, we have the mass $$m$$ to calculate stuff. So in rotational motion, I should use the moment of mass $$mr$$ in mass's place. But why do we use the second moment, which is $$mr ^2$$ ?

I have searched around in this website, but still very confused! I hope someone can help me understand it.

Inertia is something which causes the force or torque to get reduced/multiplied to give it's effect i.e. acceleration.

Torque, sometimes, is defined as the rate of change of angular momentum $$\boldsymbol{\tau} = \frac{d\mathbf{L}} {dt}$$ For explanatory purpose, let's assume that the body is going into a circular motion and we are considering a point mass at a distance of $$r$$ from the axis of rotation (the pivot point) $$\tau = \frac{d }{dt} \left(r~mv\right)$$ $$\tau = \frac{d}{dt}\left( r~ m\omega~r\right)$$ $$\tau = mr^2 \frac{d\omega}{dt}$$ Now, if we compare this to our Newton's Second Law, $$F = m \frac{dv}{dt}$$ we, at once can, see that rotational analog of translation inertia is $$mr^2$$.

You can break this myth by considering the following example:

$$\mathbf v = \boldsymbol {\omega} \times \mathbf r$$

And

$$\mathbf a = \boldsymbol {\alpha} \times \mathbf r$$

But

$$\boldsymbol {\tau}= \mathbf r \times \mathbf F$$

and

$$\boldsymbol {\ell}= \mathbf r \times \mathbf p$$

You should note the position of $$r$$ here. It is sometimes multiplied to linear quantity and sometimes to angular ones. So if there isn't any regularity in these then how can you expect it for moment of inertia?

Centre of mass is the first moment of mass , collection of point mass.

And second is moment of inertia which is collection of point masses.

In translation motion we use mass as a point(center of mass) but in rotation motion we doesn't consider mass as a point because it is distributed . Suppose a cylinder is rolling about the axis passes through center

http://hyperphysics.phy-astr.gsu.edu/hbase/ihoop.html

That axis of rotation is better for rolling than the other axis.

If you choose other axis then it is difficult to rotate. And moment of inertia of a body is not fixed it's depends on axis (we choosen) and distribution of mass(geometry)

To see why the moment of inertia contains a factor of $$r^2$$ and not just a factor of $$r$$, consider the following argument.

For a rigid body we would like an equation that relates the torque about a given axis to the angular acceleration about this axis (often we choose an axis that is fixed, but that is not essential).

Suppose there is a small particle with mass $$m$$ that is constrained to rotate about an axis $$O$$ with a constant (perpendciular) distance $$r$$ from $$O$$. If we apply a force $$F$$ to the particle then we know that

$$F=ma$$

If the angular acceleration about $$O$$ is $$\alpha$$ then $$\alpha = \frac a r$$ so $$a=r\alpha$$. The torque about $$O$$ is $$T = Fr$$ so we have

$$T = Fr = mar = (mr^2) \alpha$$

For an extended rigid body we need to integrate this equation of motion across the whole body and we get

$$T =\left (\int r^2 dm \right) \alpha = \left (\int \rho r^2 dV \right) \alpha = I \alpha$$

Mass moment of inertia arises in the calculation of angular momentum of a rigid body.

• Angular momentum is the infinite sum of moment of momentum $$\boldsymbol{L} = \sum_i \boldsymbol{r}_i \times \boldsymbol{p}_i$$

• Momentum is the scalar product of velocity $$\boldsymbol{p}_i = m_i \boldsymbol{v}_i$$.

• Velocity is the moment of rotation $$\boldsymbol{v}_i = \boldsymbol{\omega} \times \boldsymbol{r}_i = \boldsymbol{r}_i \times ( - \boldsymbol{\omega} )$$

• So angular momentum is related to the moment of moment of rotation, also known as 2nd moment of mass, or mass moment of inertia.

$$\boldsymbol{L} = \sum_i \boldsymbol{r}_i \times m_i \left( -\boldsymbol{r}_i \times \boldsymbol{\omega} \right) = \mathbf{I}\, \boldsymbol{\omega}$$

$$\mathbf{I} = \sum_i \left( -m_i [\boldsymbol{r}_i \times] [\boldsymbol{r}_i \times] \right)$$

where the cross products are converted into a matrix $$[\boldsymbol{r}_i \times] = \left[ \matrix{0 & -z_i & y_i \\ z_i & 0 & -x_i \\ -y_i & x_i & 0} \right]$$ and the combined operation yields that 2nd moment terms $$(-[\boldsymbol{r}_i \times][\boldsymbol{r}_i \times]) = \left[ \matrix{ y_i^2+z_i^2 & - x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2+y_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2} \right]$$