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Suppose there is one rod and a force F is applied in two cases:
Case 1: exactly at the middle on center of mass
Case 2: at one end of rod enter image description here

In which case would the center of mass of rod be accelerated more? Or would acceleration of its center of mass be the same in each case? I have worked towards this and found case 2 would impart a greater acceleration. I just want to check my answer and see if I am making a mistake here.(The question just came into my head and I don't have any source to check the answer from) $$My$$ $$solution:$$ In case 1 the rod would undergo a translational motion and C.O.M. would get an acceleration of $F/M$. In case 2 the rod would rotate about its other end and a torque would act which would be equal to $Fl$. Equating this to $I\alpha$ we get $\alpha$ = $3F/Ml$ and since C.O.M. is at $l/2$(rod is of uniform density) acceleration would be equal to $3F/2M$. Is my solution and thinking correct?

DeepK
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2 Answers2

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The same force applied anywhere on a rigid object causes the same translational accelleration.

The difference is that forces not applied in the direction of the center of mass will also cause some rotational accelleration.

Remember that F and A in F=mA are vectors. You can therefore treat the vectors as components in any orthagonal system you like. One way to break up F is in the direction towards the center of mass and in the plane perpendicular to that. Only the part in the plane perpendicular to the direction to the center of mass will cause roational accelleration. The entire F causes translational accelleration.

Another way to look at the applied force is to break it down into the force applied to the mass that only causes translational accelleration, and a torque that only causes rotational accelleration.

The first is simply the force divided by the mass. For the purposes of finding the positional change of the center of mass, it makes no difference where the force is applied. The center of mass will move exactly the same way in your cases 1 and 2.

Torque is the cross product of the vector from the center of mass to where the force is applied, and the force vector. That torque causes rotational accelleration, but has no effect on the position of the center of mass. In your case 1, the torque is 0 since the vector from the center of mass to the force application point is zero. If you want to say that the rod has some thickness and that therefore the force is being applied one rod radius left of the center of mass, then the torque is still zero. That is because the force vector and the vector from the center of mass to the application point are parallel, so their cross product is zero. It should also make intuitive sense that case 1 doesn't cause the rod to rotate, just accellerate to the right.

In case 2, the cross product is clearly non-zero. Let's say the rod is 1 m long and the force is 3 N to the right. The vector from the center of mass to the force application point is therefore 1/2 m up. This vector cross the force vector is 1.5 Nm into the plane of your diagram, which will cause a clockwise accelleration around the center of mass. Since this is only a torque, it has no effect on the position of the center of mass.

Olin Lathrop
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  • So would acceleration of COM be the same in each case? – DeepK Jul 16 '14 at 12:53
  • Your first two sentences would appear to contradict each other. Maybe rewrite the first to say "total translational accel" , so that it's easier to see that some of the translational is in the form of the angular momentum? E.g. I can put a ton of energy into a frisbee and it'll just hover. – Carl Witthoft Jul 16 '14 at 13:26
  • @CarlWitthoft In my experience translational excludes rotational. To be pedantic about the case of the hovering Frisbee, you have applied a torque, not a force. I do agree that the language can be confusing, but the language in the first two sentences looks ok to me. – garyp Jul 16 '14 at 14:08
  • @Carl: It makes no sense to say that some of the "translational" is in the form of any kind of momentum. We are talking about forces and accelleration here. Momentum was never mentioned. In any case "translation" specifically excludes rotational components. – Olin Lathrop Jul 16 '14 at 15:07
  • So if some of the total force applied causes rotational effects, clearly it can't cause the same translational effect (as when there's no rotation). That's what I was trying to point out – Carl Witthoft Jul 16 '14 at 15:45
  • @Carl: But it does. Your view is wrong. 1 N to the right applied anywhere on the rod will cause the same accelleration of the center of mass to the right. – Olin Lathrop Jul 16 '14 at 15:50
  • One more question- if a force is applied as in scenario 2 would the rod rotate about its center of mass or about its other end? – DeepK Jul 16 '14 at 18:15
  • @user: With no external forces, rotation of a rigid body is always about its center of mass. – Olin Lathrop Jul 16 '14 at 19:02
  • @OlinLathrop Thank you, that is where i was wrong... it is all clear now... thanks... – DeepK Jul 17 '14 at 05:51
  • @OlinLathrop I wish you would use math formatting for better readability when posting here. – John Alexiou Jul 17 '14 at 12:44
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Equations of motion explained:

  1. Sum of forces on body equals mass times acceleration of center of mass. $$\sum \vec{F} = \frac{{\rm d} }{{\rm d}t}(m\vec{v}_{cm}) = m \vec{a}_{cm}$$
  2. Sum of moments on center of mass equals mass moment of inertia times angular acceleration of body, plus velocity related terms. $$\sum \vec{M}_{cm} = \frac{{\rm d} }{{\rm d}t}(I_{cm}\vec{\omega}) = I_{cm} \vec{\alpha} + \vec{\omega}\times I_{cm} \vec{\omega}$$

In your case if the sum of forces are the same (regardless of where they are applied) the acceleration of the center of mass will be the same.

The confusion may arise from the problem of elastic impact with a rod (when there is an unknown momentum exchange and not a known force exchange) where the point of impact affects the resulting motion. See https://physics.stackexchange.com/a/29160/392 and https://physics.stackexchange.com/a/126117/392 for a similar answers.

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John Alexiou
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