# In pure translation why the body does not exhibit rotation only and only about the CM? [closed]

Suppose 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 ?

## closed as unclear what you're asking by ja72, ACuriousMind♦, user36790, CuriousOne, GertJun 15 '16 at 1:30

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• When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed-i wonder how those two forces appear in your problem? – drvrm Mar 12 '16 at 12:55
• Check Irodov's Problems in General Physics. The first one i think in dynamics of a rigid body. you'll know – Subhranil Sinha Mar 12 '16 at 14:16
• I realized I understood your question in a different way so don't mind my answer. For a more spot-on answer, you can look at this : physics.stackexchange.com/questions/93698/… – Starior Mar 12 '16 at 16:08
• You have specified the forces applied and the acceleration state. Unfortunately they are inconsistent as specified. – ja72 Jun 8 '16 at 14:10
• Possible duplicate of Do objects rotate around the torque vector or its center? – ja72 Jun 8 '16 at 14:10

• 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 7 '16 at 18:24