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

Question

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.

Answer 1

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 $mr^2$.

Answer 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?

Answer 3

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)

Answer 4

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

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

Answer 5

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 $(-[\boldsymbol{r}_i \times][\boldsymbol{r}_i \times]) = \left[ \matrix{ y_i^2+z_i^2 & - x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2+y_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2} \right]$