0
$\begingroup$

Suppose I push an object, for example a rod with two weights attached to the ends. Then, the object will move forward, but will also rotate, if the push is not at the center of mass (COM).

How is the trajectory affected, if the object itself changes such that the COM changes midflight? For example, if one weight moves smoothly to the other side?

I guess the translational movement should not change, but, if I understand this right, the rotational movement is always around the COM.

So, if I push at the COM at the start, I have no rotation. So, midflight, nothing changes and the object moves in a straight line without rotating.

But when I do not push at the COM at the start, the object has some rotation to it. And this rotation should change, right? But then, what are the exact equations for this motion? How does this change of rotation look like?

I assume there is no friction involved. Me and the object are far away in outer space.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Jan 1 at 14:54
1
$\begingroup$

Consider the following system, moving on a flat and frictionless surface.

A uniform bar has two masses $1$ and $2$. $1$ is attached to the bar (rigidly), $2$ can move freely and without friction along the bar.

We now engineer it so that at $t=0$ the CoM translates with velocity $v$ and the bar plus masses system rotates about the CoM with angular velocity $\omega$.

Rotation

Firstly, due to Galilean invariance, the rotation and translation happen completely independently from each other. And with nothing impeding translation, $v$ remains constant for all time.

However, there's no centripetal force keeping $2$ in place, so it will move away from the initial CoM. The moment of inertia of the bar plus masses will also change. In fact it will increase.

There's no friction, so rotational kinetic energy $K_r$ is conserved. At time $t=t_1$ we would have:

$$\frac12 I\omega^2=\frac12 I_1\omega_1^2\tag{1}$$

where $I$ is the initial moment of inertia. For simple systems, $I_1$ could be expressed as $f(I,\omega(t))$.

Taking the time derivative of $(1)$ and bearing in mind the $\text{LHS}$ is a constant, then:

$$\frac{\text{d}(I_1\omega_1^2)}{\text{d}t}=0$$

would be the Newtonian equation of motion for the rotational part of the motion.


As regards the notion that the CoM will move sideways, the answer to this question puts that to rest, IMO. After the system has been set in motion no external forces act on it anymore. This means that the CoG must remain in a uniform, linear motion for all of time.

$\endgroup$
8
  • $\begingroup$ So, the COM shifts to the right, following the mass $2$. Hence, the axis of rotation also changes? What is the equation for this movement, one rotation smoothly transforming into another rotation? $\endgroup$
    – StefanH
    Dec 29 '20 at 15:53
  • $\begingroup$ No, the axis of rotation doesn't change. It's a common misconception that rotation is only possible about the axis through the CoM but that is not true. The CoM doesn't move 'sideways' because that would require kinetic energy. $\endgroup$
    – Gert
    Dec 29 '20 at 15:56
  • $\begingroup$ Ok, but, if an object gets an initial push, the rotation is always around the COM? But this may change if the distribution of mass in the object changes. $\endgroup$
    – StefanH
    Dec 29 '20 at 16:05
  • $\begingroup$ Initially yes, the rotation is about the axis through the CoM. $\endgroup$
    – Gert
    Dec 29 '20 at 16:08
  • $\begingroup$ physics.stackexchange.com/questions/603872/… this very new question is related and worth contemplating. $\endgroup$
    – Gert
    Dec 29 '20 at 18:05
1
$\begingroup$

First let's take the situation when the object is moving without rotation, and by some internal mechanism, one of the weights moves.

For an inertial observer, by conservation of momentum, the other weight (together with the rod that connects both weights) must also move, keeping the COM in the same previous straight line.

If the same happens for a rotating object, both weights also move to keep the COM in the same straight line.

About the rotating, in the most general case for a rigid body, we can only say that the angular momentum is constant, but the magnitude and direction of rotation may change.

The equation for a rigid body is: $\mathbf L = I\omega$

where: $\mathbf L$ is the angular momentum, $\omega$ is the vector angular velocity ($\omega_x$ , $\omega_y$ ,$\omega_z$) and $I = \int_v \rho M dv$ is the inertia matrix. $M$ is the matrix:

\begin{Bmatrix} (y^2 + z^2) & -xy & -xz \\ –yx & (z^2 + x^2) & -yz \\ -zx & –zy & (x^2 + y^2) \end{Bmatrix}

But if one of the balls is free to move back and forth along the rod, it is no more a rigid body. The EOM is still more complicated.

$\endgroup$
2
  • $\begingroup$ What you say in the first three paragraphs implies that the trajectory of the COM is not a straight line anymore, right? $\endgroup$
    – StefanH
    Dec 30 '20 at 14:20
  • $\begingroup$ It is a straight line. We imagine that only one ball is free. But the situation is symmetric. Both (the free ball and the rest of the body) are equally free to move with respect to the other. $\endgroup$ Dec 30 '20 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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