What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass?

Question

Say I have a rigid body in space. I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.

What is the proof of this?

Answer 1

A very simple reason would be that if the body rotated about some point other than the center of mass, the center of mass in the ground frame would be in circular motion.

Now we know that the motion of the center of mass is governed by EXTERNAL FORCES ONLY, and in case of a force applied for a short time, there is no external force acting on the center of mass subsequently.

So we can say that the subsequent motion of the center of mass will be linear(and not circular , which it would be if the body rotated about some other point). And as the body has some angular momentum, it will rotate about the center of mass!

Answer 2

What you are talking about is called the instant center of percussion. To purely rotate a rigid body about an axis (the rotation axis) a force needs to be applied along the axis of percussion which is a) perpendicular to the rotation axis, b) on the far side of the center of gravity from the pivot and c) located a distance $ \ell =c + \frac{I}{m c}$ from the pivot ($m$ mass, $I$ mass moment of inertia about cm and $c$ distance between pivot and cm).

Derivation

Consider a body with desired rotation $ \vec{\omega} = (0,0, \omega_z)$ about a point A aligned with a local $\hat k$ axis, and the center of gravity located along the local $\hat i$ axis, with coordinates $\vec{c} = (c_x,0,0)$.

An impulse with components $\vec{J}=(J_x,J_y,J_z)$ is applied at a location $\vec\ell = (l_x,l_y,l_z)$ relative to A with the equations of motion at the center of mass

$$ \vec{J} = m \left( \hat 0 + \vec{\omega} \times \vec{c} \right) \\ (\vec{\ell} -\vec{c} ) \times \vec{J} = I \vec{\omega} $$

in components the above is

$$ \begin{pmatrix} J_x \\ J_y \\ J_z \end{pmatrix} = m \begin{pmatrix} 0 \\ 0 \\ \omega_z \end{pmatrix} \times \begin{pmatrix} c_x \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ m c_x \omega_z \\ 0 \end{pmatrix} $$

So $J_x=J_z=0$ making $\vec{J}$ to be along the local $\hat{j}$ axis.

$$ \begin{pmatrix} \ell_x - c_x\\ \ell_y \\ \ell_z \end{pmatrix} \times \begin{pmatrix} 0 \\ J_y \\ 0 \end{pmatrix} = \begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \end{bmatrix} \begin{pmatrix} 0 \\ 0 \\ \omega_z \end{pmatrix} $$

$$\begin{pmatrix} -(m c_x \omega_z) \ell_z \\ 0 \\ (m c_x \omega_z) (\ell_x-c_x) \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ I_z \omega_z \\ \end{pmatrix}$$

with solution $\ell_z =0$ and $\boxed{\ell_x = c_x + \frac{I_z}{m c_x}}$. Note that the value of $\ell_y$ is irrelevant since it along the force axis $\vec{J}$.

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Here are some reference posts:

See relevant answer to a similar question (https://physics.stackexchange.com/a/81078/392)

The full equations of motion about an arbitrary point are derived in (https://physics.stackexchange.com/a/80449/392)

Answer 3

I've read that if I during some short time interval apply a force on the body at some point which is not in line with the center of mass, it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass.

To my understanding, your question is flawed. If a single force is applied to a rigid body under the influence of no other forces, either:

1. The line of action of the force passes through the center of mass, causing a pure translation and no rotation
2. The line of action of force does not pass through the center of mass, in which case you end up with a pure rotation about an axis which does not pass through the center of mass. In other words the instantaneous axis of zero velocity induced by a single force can never be the center of mass.

If you apply an eccentric force, the center of mass of the body will undergo a linear acceleration, and the body itself will undergo an angular acceleration. In a fixed reference frame, this can be viewed as a pure rotation about a certain point, but this point will never be the center of mass of the body.

Answer 4

What is the proof of this?

The proof of Chasles’s Theorem relating to the rotational and translational displacements of a rigid body is done in very many texts including Appendix 20A of this document produced by MIT.

In essence this shows that, if a force whose line of action does not pass through the centre of mass of a rigid body, the applied force is equivalent to the same magnitude and direction force which passes through the centre of mass of the body which produces only translation acceleration of the centre of mass of the body and a couple which produces only rotational acceleration of the body.

The actual motion of the body is dictated by the sum of these two accelerations.

Suppose a force $\vec F$ is applied to a body whose line of action does not pass through the centre of mass $C$ of the body.

Adding two forces $\vec F_1$ and $\vec F_2$ at the centre of mass of the body such that $\vec F = \vec F_2$ and $\vec F_1 + \vec F_2 =0$ as shown in the diagram below.

There is now a force $\vec F_2 (= \vec F)$ acting along a line through the centre of mass of the body, $C$, which will only produce a translational acceleration of the body and a couple consisting of the two forces $\vec F$ and $\vec F_1$ and of magnitude $Fd$ which will only produce a rotational acceleration of the body.

Answer 5

Assume a very small particle embedded in the Rigid body of mass $m$. Let us find out its Torque or moment of force $\vec{\tau}$ about an arbitrary point $p$.

$\vec{\tau} = \vec{f} \times \vec{r}$

where $\vec{r}$ is a displacement of this particle from point $p$.

The total Torque on the rigid body will be some of $\tau$ of all the particles. If this $\tau$ has a non-zero value then the body will be rotating.

Lets find out the total Torque, $\Gamma$

$\Gamma = \Sigma{\tau}$

$\Rightarrow \Gamma = \Sigma{ \vec{f} \times \vec{r}}$

$ \Rightarrow \Gamma = \Sigma{ m \, \vec{a} \times \vec{r}}$

As The body is said to be rigid, therefore all the points on this body will be having same accelerations at ever instance. Also, Cross product is distributive ref, therefore, we can take $\vec{a}$ out of summation.

$\Rightarrow \Gamma = \vec{a} \times \Sigma{ m \, \vec{r}}$

now, if point $p$ is center of mass then, $\Sigma{ m \, \vec{r}}$ is zero. ref

Therefore, $\Gamma$ is zero and rigid body will not rotate at all.

NOTE: $\times$ is the vector cross product operator.

Answer 6

I think the point is that in free space linear and angular momentum are both separately conserved quantities. (This is implied so long as your space has translational and rotational symmetry.) If the total linear momentum of the rigid body is always constant after the push, then the center of mass must be moving in a straight line, at constant velocity. From this it follows that the rotation must be about the center of mass.

Answer 7

You can find the change in angular momentum of a rigid body by simply evaluating it: $\frac{d\mathbf{L}_{total}}{dt} = \sum_p m_p\left(\mathbf{R}+\mathbf{r}_p\right) \times \frac{d}{dt}\left(\mathbf{V}+\mathbf{v}_p\right)\,$

$\frac{d\mathbf{L}_{total}}{dt}= M \mathbf{R} \times \frac{d\mathbf{V}}{dt} + \sum_p m_p \mathbf{r}_p\times \frac{d\mathbf{v}_p}{dt}$ Here I've broken up the position of the $p^{th}$ component into a center of mass part and a relative part.

Note that $m_p \frac{d\mathbf{v}_p}{dt}$ is precisely the force acting on one part of the body. You can show that internal forces (forces between particles) don't contribute to the torque (basically due to them being equal and opposite, so they cancel when you sum), so only external forces are important.

Only the component of that force that is perpendicular to $\vec{r}_p$ survives the cross product, and sets the body rotating. In other words, the statement " it would start rotating about an axis which is perpendicular to the force and which goes through the center of mass" is a property of the cross product in that equation. Why about the center of mass? Well, you can evaluate the angular momentum about any line (more precisely, in any plane), and it neatly factors into a part that is the motion of the COM about that axis, and a part that is the motion of the body about the COM. If you choose the axis to go through the COM then the first part vanishes by the cross product. Anyway, the calculation above factors in the same way, as you can see.

You can check out the nitty gritty here which I typed up a long time ago. Hopefully it isn't too confusing.

Cheers

Answer 8

One can make reasonable assumptions to investigate the problem in a simple manner. Here is my reasoning about this question.

For the sake of simplicity let us assume we have a spherical object of radius R in the outer space. Let there be a hook at the surface of the sphere from which we can attach a string. Imagine we are equipped with a rocket system that can give us momentum to move about.

Now, we hold one end of the string and move away from the sphere in a direction that the string, when it becomes taut, is not parallel to the radius of the sphere. The force we exert on the sphere in that direction can be analysed into the tangent and the perpendicular to the surface of the sphere. If $\theta$ is the angle between the string and the normal to the sphere we have:

Tangent component: $F_T=F\sin(\theta)$

Normal component: $F_N=F\cos(\theta)$.

The normal component is parallel to the radius of the sphere and passes through the centre (CM) and has no moment. This component will pull the sphere in the normal direction.

The tangent component has a moment with respect to the centre

$M=FR\sin(\theta)$.

This component would rotate the sphere, should the axis of the sphere be pivoted, but it is not! However, I believe that, due to the inertia of the mass of the sphere, it would be sufficient to give pivotal leverage for the tangent force to rotate the sphere. The law of conservation of energy must be written, for a short time interval of application of the force, in the form

${\bf {F.x}} = {\frac {1}{2}}mv^2+ {\frac {1}{2}}I{\omega}^2 $

where:$\bf x$ is the displacement of the sphere, while the first term on the RHS is the kinetic energy due to the linear motion, and the second is the kinetic energy due to the rotational motion. Note that, as the sphere has no fixed axis, it will rotate about the axis which is peprepndicular to the great circle passing through the point of the hook, and the $F_T$ is tangent to it. Hence the axis will be perpendicular to $F_T$ and $F_N$ and so it is perpendicular to the force $\bf F$. This will be the case for any direction of $\bf F$.

Why should the axis of rotation pass through the CM? The poitn here is that the object is rotating freely. Is not constrained to rotate about an arbitrary axis. Without going into mathematics, a quick argument from physics point of view is that, if the axis passed through another point, the rotational motion would be unstable. I mean that for a freely rotating object, there is a minimum state of energy, and this is when the axis of rotation passes through the CM. If it passed through some other point, then according to the parallel axis theorem, the inertia of the object would be higher, hence higher energy of the system. It is like you bring an object at a certain height near the surface of the earth and then you set it free. It will fall to the lowest energy state, and that is when it is on the ground.