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enter image description hereSuppose I have a rod in pure translation as shown in the picture with some constant acceleration. The two forces $F_1$ and $F_2$ act as shown. So clearly due to the definition of pure translation every point in the rod has the same velocity (say $\vec v$) so relative velocity of any point w.r.t to another point on the rod must be zero.

But, if I consider the net torque about $A$(or $O$), clearly, there is a non-zero torque due to $F_2$ ( or $F_1$ if you consider $O$) so clearly there must be some angular acceleration about that point. which would mean that the there is some non-zero relative velocity of other points on the rod about $A$(or $O$)

I don't understand what I'm getting wrong here.

And from what I read the rod must have no rotational motion only and only about the Centre of Mass. why is so ?

Please please explain to me clearly.

1 Answers1

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In this case, the rod cannot experience only pure translation. Because resultant torque acting on the rod isn’t equal to zero.

If acting points of the forces are fixed, the rod will rotate counter clockwise until it is parallel with the forces.

enter image description here

So, until that final state of the rod is established (without damping forces, this state will never be established), the rod will experience both of rotational and translational motions.

lucas
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  • Without damping forces the oscillations would continue indefinitely – Rick Jun 07 '16 at 17:56
  • I think the oscillations will be removed finally even without damping forces. – lucas Jun 07 '16 at 18:08
  • They would not... the torque about the COM would be exactly proportional to the sine of the angle from horizontal. The corresponding differential equation would then be: $\ddot \theta = k sin(\theta)$ which is equivalent to an ideal (undamped) pendulum which will oscillate forever. – Rick Jun 07 '16 at 18:24