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Q:If we know that the acceleration vectors of all points of the rigid body are equal to each other at all points of time, can we conclude that the body is undergoing pure translational motion?

I recently learnt that both the necessary and sufficient condition for a Rigid body to be in pure translational motion is that, “All the points of the rigid body should have the same velocity vector at any instant of time.” The velocity may change, but the velocities(vector) of all the points are equal to each other at every instant.

From this we can prove that that the acceleration vectors are equal as well (i.e. it’s a necessary condition)

But is that a sufficient condition as well?

I tried to come up with an example where this statement isn’t true, but unsuccessful...

So can someone help me prove or disprove this statement?

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  • $\begingroup$ What if all points have acceleration equal and in the x-direction? then you have linear acceleration and no rotational acceleration. I suspect you need to rewrite to say "angular acceleration" and "angular velocity" $\endgroup$ Commented Nov 23, 2020 at 13:55
  • $\begingroup$ A circular ring of infinitesimal thickness rotating at a uniform angular speed, as a counterexample, has all the points on it accelerating toward the center of the ring -- their centripetal acceleration vectors are not 'equal' in direction. $\endgroup$
    – Yejus
    Commented Nov 23, 2020 at 14:54
  • $\begingroup$ Sorry about that, I typed rotational instead of translational. I edited the question now. $\endgroup$
    – Mysterio
    Commented Nov 23, 2020 at 14:55

3 Answers 3

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This definition should help. "A rigid body moves in pure translation if each particle of the body undergoes the same displacement as every other particle in any given time interval" [Halliday and Resnick, Physics].

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If all points of the body have the same acceleration, we can say for any pair of them: $$\frac{d\mathbf v_i}{dt} = \mathbf k$$ $$\frac{d\mathbf v_j}{dt} = \mathbf k$$

Here $\mathbf k$ can be a function of $t$.

Integrating: $\mathbf v_i - \mathbf v_j = \int (\mathbf k - \mathbf k) dt = 0$

So, all points have the same instant velocity as well.

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Q:If we know that the acceleration vectors of all points of the rigid body are equal at all points of time, can we conclude that the body is undergoing pure rotational motion?

No it is neither necessary nor sufficient that the acceleration (not the vector, but it's magnitude) being constant implies that the motion is purely rotational, because for pure rotational motion, the acceleration must also be directed to the center of rotation. On the other hand, if the acceleration vector were to be constant, then the motion isn't circular either, rather the motion could either be linear or parabolic.

So to conclude that the body is performing purely rotational motion, you will have to prove that the magnitude of acceleration is constant and is always directed towards a fixed point in space. This is the necessary and sufficient condition for a body to perform pure rotational motion.

It is not a necessary or a sufficient condition for linear translation motion, as constant acceleration motion could be parabolic (motion in uniform gravitational field) as well, and non constant acceleration can also produce linear motion (example, shm due to a spring)

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  • $\begingroup$ Really sorry about that, I typed rotational instead of translational. I edited the question now. $\endgroup$
    – Mysterio
    Commented Nov 23, 2020 at 14:57
  • $\begingroup$ Magnitude of acceleration can't be constant: $a_r = \omega^2r$ The farther from the point, the larger the radial acceleration for a given angular speed. $\endgroup$
    – Bill N
    Commented Nov 23, 2020 at 15:50
  • $\begingroup$ @Mysterio again it isn't sufficient or necessary either, the motion could be parabolic with constant acceleration, and it isn't necessary that a straight line motion have a constant acceleration $\endgroup$
    – SK Dash
    Commented Nov 23, 2020 at 15:56
  • $\begingroup$ Hmm I think the question is slightly misleading, what I meant to say was that at a certain instant, all the velocity vectors are equal to each other, and may change with time, but always “equal to each other” $\endgroup$
    – Mysterio
    Commented Nov 23, 2020 at 16:06
  • $\begingroup$ @Mysterio I believe there is only one velocity and that is of the body in consideration, if you're talking about the rotation of a rigid body about its axis, even then, the velocity vectors at any point of the body are never equal, rather at the same distance from the sphere, the speeds of the points are equal. The condition given above for pure rotational motion, applied at all the circles, would be necessary and sufficient to prove that the rigid body is performing pure rotation about some axis $\endgroup$
    – SK Dash
    Commented Nov 24, 2020 at 0:31

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