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If a rod is on a frictionless plane, and a force is applied on one of it's end, will there be both, translation + rotation motion? Also, if only a single force is applied on a body that does not pass through Centre of Mass, will it always produce rotation + translation?

Sachin Chaudhary
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2 Answers2

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Suppose that you have a force $\vec F$ acting at a point $A$ on the rod as shown in the diagram below.

enter image description here

Add two forces of equal magnitude but opposite in direction whose line of action is parallel to the original force but with those forces acting at the centre of mas.

This results in a force $F$ (red) acting through the centre of mass which provides the translation acceleration of the centre of mass and two forces $\vec F$ and $-\vec F$ (grey) which constitute a clockwise couple of magnitude $Fd$ which will produce the angular acceleration of the rod.

So if the line of action of the applied force does not go through the centre of mass of the rod you will always get a translational acceleration and a rotational acceleration.

Farcher
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  • Thank you. I can see the explanation for rotational acceleration in your answer, but where the translation acceleration could be seen from here? – Damir Tenishev Dec 15 '22 at 17:55
  • @DamirTenishev The grey forces produce the torque and hence the rotation and the red force which acts through the centre of mass produces the linear acceleration. – Farcher Dec 15 '22 at 18:25
  • The exact question is, what makes you think that these grey forces won't balance this red force in linear acceleration term? Hint: if A is infinitely far from C we have exactly this: only rotation will occur. The question is how to prove that with a finite distance between A and C this won't happen. Another concern, with your explanation it sounds like there is no difference where to apply the force; the resulting linear force/acceleration will be the same. This sounds a little bit odd, life experience tells me that kick to center much more effective. – Damir Tenishev Dec 16 '22 at 01:53
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    life experience tells me that kick to center much more effective this is due to frictional forces being present. Look at Appendix 20A Chasles’s Theorem and note that many things in rotational dynamics are counterintuitive. – Farcher Dec 16 '22 at 08:43
  • Thank you so much for the link, this is just what I needed. It seems that you are experienced in the subject, could you please consider my question Rotation of a system of bodies on an axis and the root cause for it Rotation of the systems of two bodies connected by a motor? Maybe you can help me to figure out what is a missing part in my understanding of rotation. – Damir Tenishev Dec 16 '22 at 14:27
  • @Farcher Are your forces instant(just on hit) or continuous forces? – user707264 Oct 11 '23 at 12:17
  • The diagram is for an instant of time because the rod will rotate and so its orientation relative to the force would change. – Farcher Oct 11 '23 at 16:41
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Simple answer is that in a frictionless environment force will act on the center of mass to produce a linear acceleration. If the force is acting away from the center of mass ie off set and therefor forming a moment ( m x distance from the center of mass) then the acceleration will be angular and centered on the center of mass ie the rod will rotate in the direction of the applied force, around its center of mass

Spanners
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