1
$\begingroup$

There are two different sorts of inertia: inertial mass and moment of inertia.

I am currently reading about moment of inertia. Now, I know inertia is an important concept; with it, we can determine how difficult something is to move. For linear motion, the difficulty of altering an object's position is dependent upon the mass of the object being moved. For rotational motion, the difficulty in rotating an object around an axis is dependent upon the mass, and the radius of that mass to the the axis of rotation?

  1. Why do we need to different measures of inertia?

  2. And for rotational motion, why is the inertia dependent upon mass, and its radial distance from the axis of rotation?

$\endgroup$
1
  • 1
    $\begingroup$ The "inertia" in both the cases is the same thing. You'll want to look into the general use of the term "moment" and understand that "moment of inertia" is short for "the first moment of inertia". $\endgroup$ Nov 25, 2012 at 19:13

1 Answer 1

4
$\begingroup$

The moment of inertia is merely a generalisation/application of the ‘usual’ inertia to rotations. Since translations and rotations are different kinds of motion, it appears sensible (to me) to have different kinds of inertia associated with them.

Regarding your second question: Imagine a particle at position $(x,0,0)$ which you would like to rotate with angular velocity $\omega$ about the $(0,0,z)$ axis. To do so, you have to initially accelerate the particle along the $(0,y,0)$ axis to velocity $v_y = \omega x, v_{x,z} = 0$, as this is the velocity the particle would have at this point if it were already rotating.

As you can clearly see, the momentum $p$ associated with this velocity is proportional to $r$ ($p_y = m v_y = m \omega x$), hence it takes more energy to accelerate a particle to angular velocity $\omega$ if it is further away from the centre of rotation. As this is exactly the quantity described by the ‘moment of inertia’, the moment of inertia depends on the radial distance of the mass.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.