# Will mass distribution affect the motion of a rigid object?

Assuming I have two rigid objects sitting at rest in a 2D plane with some friction coefficient. Both objects have the same shape, same total mass, same center of mass but DIFFERENT mass distribution. When applying the same force on both objects, will they always behave the same?

My intuition tells me the moment of inertia will only depend on center of mass rather than mass distribution, am I right?

My intuition tells me the moment of inertia will only depend on center of mass rather than mass distribution, am I right?

Nope.

Moment of Inertia depends on both the mass and the distribution of the mass. Further away from the axis of rotation, a unit of mass will cause a greater moment of inertia. Therefore, if you have two bodies of the same mass, same centre of mass, but a different distribution; they can have very different moments of inertia. Consider the difference between a circular plate with even mass distribution, and a circular plate with mass concentrated at the edges.

To simplify, you can consider the objects as a sum of point masses and their moments of inertia. Each point mass has a inertia given by $I = mr^2$ where $m$ is the mass of the point; and $r$ is it's distance from the axis of rotation. You should be able to see that if most of the mass is concentrated far away; the $mr^2$ products would be bigger, leading to a greater total inertia (for $N$ point masses, the total inertia is given by $I_{net} = \sum_{i = 1}^N m_N \ r_{N}^{2}$).

If you are interested in computing the rotational motion upon applying the force, you need to consider rotational inertia (i.e., moment of inertia). The rotational inertia is a function of mass distribution. See Moment of Inertia and notice the $r^2$ inside the integral.

There are two parts to the motion of a rigid body. The motion of the center of mass (point) and the rotation about the center of mass.

To describe the path of the center of mass, only the forces and mass is needed $$\sum \mathbf{F}_i = m \,\mathbf{a}_C$$ Where: $\mathbf{F}_i$ is a force acting on body, $\mathbf{a}_C$ acceleration of the center of mass

But to describe the rotation you need the mass moment of inertia and the location of the forces. The mass moment of inertia depends on the distribution of mass. In 2D

$$\sum \| (\mathbf{r}_i - \mathbf{r}_C)\times\mathbf{F}_i \| = I_C \alpha$$

Where: $\mathbf{r}_i$ is the location of force $\mathbf{F}_i$, $\mathbf{r}_C$ the location of the center of mass, $\alpha$ the rotational acceleration and $I_C$ the mass moment of inertia about the center of mass.

The definition of the mass moment of inertia in 2D is

$$I_C = \iint (x^2+y^2) {\rm d}m$$ for each small mass ${\rm d}m$ located at $(x,y)$ relative to the center of mass.

Your intuition is incorrect. A simple example is a pair of spheres with the same mass $m$ and radius $r$ rolling from rest down an inclined ramp of height $h$, with one of the spheres being a uniform density solid and the other hollow. At the bottom of the ramp, both spheres will have a kinetic energy $\frac 1 2 v^2 + \frac 1 2 I \omega^2 = mgh$, where $v$ is the velocity of the center of mass, $I$ is the moment of inertia, and $\omega$ is the angular velocity. Since the spheres roll without slipping, $\omega = v/r$. The velocity at the bottom of the ramp is thus $v = \sqrt{(2ghmr^2)/(I+mr^2)}$.

The moment of inertia of the solid sphere is $I_\text{solid} = \frac 2 5 mr^2$ while for the hollow sphere it is $I_\text{hollow} = \frac 2 3 mr^2$. The solid sphere will be going faster than the hollow sphere at the bottom of the ramp, and will reach the bottom of the ramp in less time than it takes the hollow sphere to reach the bottom.

## protected by Qmechanic♦Jan 3 '18 at 16:07

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