It is said in my textbook that to set a puck into both translational and rotational motion, we can simultaneously push the puck (with any linear velocity) and rotate it (with any angular velocity).

I don't think we can set an object into both translational and rotational motion with only 1 force. It requires 2 forces - one applied at the puck's center of mass, giving it linear motion; the other applied at any other points of the puck except for its cm, to set it into rotation around its cm.

However, when I tried out flicking a wite-out at its edge in real life, it not only rotates but also moves forward (ie has linear motion).

I am wondering why this happens, does it mean we only need one force to set an object into both translational and rotational motion theoretically as well?

  • $\begingroup$ See e.g. youtu.be/6dG9hb3_blo (slow motion golf balls). This is not the best example, the best examples come from billiards where spin control is a gigantic part of the game and your technique, but I think it's less arguable that the ground is playing a crucial role. $\endgroup$
    – CR Drost
    Jul 16, 2021 at 15:40
  • $\begingroup$ Any force whose line of action does not go through the center of mass is going to cause both translation and rotation. $\endgroup$ Jul 17, 2021 at 2:55

5 Answers 5


You only need one force, applied off the object's center of mass. Such a force produces rotational motion, as you know, but it must also produce linear motion. You deliver some amount of linear momentum to the object over the time you are applying the force (because that is what forces do). If there is no other force on the object that can take that momentum away, the center of mass must start moving.


Any net force that is not directly in line with the COM of an object, that doesn't have a fixed axis, will cause both spatial translation and rotation of the object. For the object to have translation only, the force must be directly in line with the COM. To have rotation only on the object, equal and opposite forces would have to be applied on equally opposing sides of the COM so that their opposing directions cancel the translation but work on opposing points to create torque.


You can do it trivially. Imagine a point mass $m$ at $\vec r_0$. At rests. Now apply a force $\vec F$ for time $\Delta t$. You can shrink $\Delta t\rightarrow 0$ and keep the impulse fixed:

$$\Delta \vec{p} = \vec F \times \Delta t$$

The particle then has momentum $\vec p$ and velocity $\vec v$. It has also acquired angular velocity about the origin:

$$ \vec{\omega} = \vec r_0\times\vec v$$

from the torque:

$$ \vec{\tau} = \vec r_0\times\vec F$$

So that's trivial, and perhaps unsatisfying. The simplest non-trivial case is an and ideal dumbbell: two equal masses $m$, on the end of a massless, rigid, stick of length $L$.

Place that at the origin along the $y$-axis. It has a moment of inertia:

$$ I = \frac 1 2 mL^2\left[\begin{array}{ccc}1&0&0\\0&0&0\\0&0&1\end{array}\right]$$

Now apply the same impulse from the trivial example to the lower mass, with $\vec F$ along the $+x$-axis. You will get COM motion parallel to $x$, and rotation parallel to $z$.


A single force (or single impulse) can cause rotation but not all rotations are possible.


In the example above a force applied on the rim of a puck introduces an overturning moment about the center of mass. This moment is calculated as $$\boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{F}$$

where $\times$ is the vector cross product. This overturning moment is responsible for the rotation of the object as it changes the angular momentum of the object, just as the force changes the translational momentum.

On the right figure, the body acquires translational momentum $\boldsymbol{p}$ (and corresponding velocity $\boldsymbol{v}$ of the center of mass) and rotational momentum $\boldsymbol{L}$ (and corresponding rotational velocity $\boldsymbol{\omega}$ about the center of mass).

$$ \begin{aligned} \boldsymbol{F} & = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{p} = m \boldsymbol{a} \\ \boldsymbol{r} \times \boldsymbol{F} & = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{L} = \mathrm{I} \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathrm{I} \boldsymbol{\omega} \end{aligned}$$

But because of the cross product, no component of this moment can be parallel to the force applied. So from the possible 3 directions of moments, only 2 can be realized as the equipollent moment of a force.

To fully prescribe the motion of a body with a single impulse, you need not only a force in a single direction but also a parallel force couple acting to provide for rotation about the axis of the force.


There are lots of arguments showing that you can:

  1. Conservation of linear momentum. Suppose you throw a glove at a puck. When the glove hits the puck, it will slow down, losing linear momentum. That momentum must go to the puck.

  2. Consider a two-engine plane. Clearly the two engines together can impart linear momentum, even though they are not applying that force through the center of mass. If each engine were not imparting linear momentum by itself, how could the two of them together do so?

  3. Consider an object as being several connected together. Suppose you have two pucks tied together, and you apply a force that goes through the center of mass of one of them, but not through the center of mass of the two-puck system. The force transmits linear momentum to the one puck, and that momentum doesn't disappear just because the puck is tied to another one.

  4. Suppose you have a rocket in orbit around the sun, and the engines are pointing along its orbit. Firing the engines will change the rocket's linear momentum, but it will also change its angular momentum around the sun.

A force of F applied over a time t imparts momentum of Ft, regardless of whether it is applied through the center of mass.


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