What I'm trying to understand

I want to understand what is the linear velocity and angular velocity result of a cube pushed by an instant force at location in the void, with no frictions, no gravity and no other forces, and total freedom.

The instant forces is applied at the top of cube and is orthogonal to the COM, as showed in the image below.

enter image description here

Tests and searches I did

Test 1 (real world)

Initially I've did this test, where I've pushed a cube with a finger in the corner. The box got either linear velocity and angular velocity. Since this test had too much bias because the gravity force that was acting, the surface was producing friction forces (though I've selected a low friction surface), and the air that was producing friction (though if low).

For These reason I've speculated that the box gained linear velocity because the ground friction, air friction, they gave back some normal forces that applied traslation movement.

Searching phase

So I not consider the test I've did before, and I've searched on the internet where I've found a forum's thread where a person says that the COM tries to not traslate when a box is pushed by a force that is not al COM, so it rotates.

Test 2 (Software)

So at this point I've two different view of the same problem, so I've thought to reproduce this problem using a software PhysX.

When the simulation software, that I've set, was started, the box started to rotate and traslates like the first test I've done.

After some checks I did on the software to be sure no forces were acting, I've understand that the definition I've readed on the forum was not correct.


I don't know how to write better than this but what I'm trying to understand is:

If is correct that the box will rotate and traslate when pushed in the side like shown in the image, what is the result linear and angular velocity? and how it's calculated?

In other word I'm searching how to calculate the linear acceleration and the angular acceleration by knowing the Force vector and the Location where the force is applied.

So sorry if I was too vague, I hope now my question is well formed and well argue to get complete response.

  • 1
    $\begingroup$ Please explain me what I'm doing wrong, instead of leave useless down votes $\endgroup$ Apr 1, 2017 at 8:31
  • 1
    $\begingroup$ You probably got that downvote because this question in its current form is not useful for this site. It looks like a homework question and there are things you need to do to get help with homework here. Please see How do I ask homework questions on Physics Stack Exchange?. $\endgroup$
    – PM 2Ring
    Apr 1, 2017 at 9:32
  • $\begingroup$ Thanks for your response, but I'm asking this to understand how a box will move when pushed by a force not applied at COM, it's not an homework. How do I have to write this question to understand this? $\endgroup$ Apr 1, 2017 at 9:43
  • $\begingroup$ If I reformulate this question by adding more information, and what I noticed I can repost it? $\endgroup$ Apr 1, 2017 at 9:46
  • 1
    $\begingroup$ I think we don't understand where your understanding is breaking down. Is the problem that you have not been exposed to how Newton's laws, or that you don't understand how Newton's laws are applied to extended objects. $\endgroup$
    – garyp
    Apr 1, 2017 at 13:55

3 Answers 3


You can use this to calculate the change in translational velocity ($\Delta v$) of the CM: Note the change in translational velocity is in the direction of the force: $$\frac{dv}{dt} = a = \frac{F}{m}$$ $$\int dv = \int\frac{F}{m}dt$$ $$\Delta v = \frac{F}{m} \Delta t$$ $\Delta t$ is the small amount of time that the force acted on the object.

Usually, specify $\Delta t$ to be small and $F$ to be large.

Also, for the rotational motion: $$\frac{d\omega}{dt} = \alpha = \frac{F*r\sin\theta}{I}$$ $$\int d\omega = \frac{F*r\sin\theta}{I}\int dt$$ $$\Delta\omega = \frac{F*r\sin\theta}{I}\Delta t$$ Where $\Delta\omega$ is the change in angular velocity,
$I$ is the moment of inertia of the object
$r\sin\theta$ is the moment arm of the force with respect to the CM.

Above are formulas for calculating the change in velocity and angular velocity.
Note that the formula uses $\Delta t$ which doesn't look like instantaneous, but you can treat it like Impulse: $\Delta p= F*\Delta t$ which can be assumed as instantaneous and produces the same results (in the change in momentum) as slow-cooked impacts.

Yes it will translate (even if there is no gravity)
and rotate (If the line of force does not pass through the center of mass, that is, if $\sin \theta$ is nonzero).

  • $\begingroup$ Ohh perfect, so the force applied at point gives a linear velocity like if it's applied on the COM without any loss. Thanks you, for your precise response this is the information that I was searching! Thanks alot!! $\endgroup$ Apr 1, 2017 at 18:30
  • $\begingroup$ By your formula above, change in translational velocity doesn't depend on the leverage between CoM and point of force application. Are you saying that the same force applied to the CoM and to the side of the box (tangentially, with a leverage) will result in the same translational speed of CoM? This sounds counter-intuitive, please confirm. $\endgroup$ Dec 15, 2022 at 18:13

By adding two extra forces both equal in magnitude to the instant force but opposite in direction you create a couple (blue) and a force (red) acting at the centre of mass.

enter image description here


As per diagram, and as far as I could gauge from the question and the comments after it, the box will rotate and translate.


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