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For a rigid body undergoing planar motion I can associate to it the concept of Instantaneous Axis of rotation $IAOR$, the point from which I can see the body undergoing pure rotation. Consider any point on the rigid body.

Here is my concern. Why is the radius of curvature of the point's trajectory and the it's distance from the $IAOR$ not the same? Take for example a rolling disc on a smooth horizontal surface.

The radius of curvature of the top most point is $4r$ but the distance of that point from the axis of rotation is $2r$ (Here the $IAOR$ is the bottom most point of the disc). What I think is that it's distance of the point from $IAOR$ should be the same as the radius of curvature. Where am I wrong?

EDIT Here's how I calculated the radius of curvature of the topmost point. It's velocity $w.r.t.$ ground is $2v$, so according to the equation of kinematics centripetal acceleration of a particle whose trajectory is known beforehand would be $R_c=\frac{(2V_0)^{2}}{a_c}$. The centripetal acceleration would remain the same$=\omega^{2}r$ or $\frac{V_0 ^2}{R}$ Plugging in the values we get the radius of curvature $=4R$ ( where $R$ is the radius of the rolling body).

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Let us start by setting some parameters.

The body is rotating with angular speed $\omega$ and its centre of mass is moving translationally with velocity $\omega r$.

Centripetal acceleration of the uppermost point about centre will be given by,

$a_c=\omega^2r $ $\tag 1$

For a purely rolling body, point of contact will be the instantaneous axis of rotation. The angular speed of the rigid body will still be $\omega$ about this axis. (You can prove it by dividing the velocity of uppermost point with respect to the $IAOR$ which will be $2v$ and its the distance which will be $2r$, and that will leave you with $\omega$.)

About this axis, centripetal acceleration of the upper most point will be given by,

$A_c=\omega^{2} (2r)$

$\therefore \space $ $A_c=2a_c$ $\tag 2$

The centripetal acceleration of the uppermost point about $IAOR$ can also be given by,

$A_c=\frac{v^2}{R'}$, where $R'$ is the radius of curvature.

$R'=\frac{v^2}{A_c}$

For a purely rolling body, velocity of the uppermost point is $2\omega r$ and zero of the $POC$.

From equations $(1)$ and $(2)$,

$R'=\frac{4\omega^2r^2}{2\omega^2 r}$,

$\therefore \space$ $R'=2r$

I thing the mistake you were doing is assuming the centripetal acceleration of the uppermost point about the $IAOR$ to be equal to $\omega^2 r$, but actually it is $2\omega^2 r$.

Now this kinda makes sense doesn't it ? Let's try doing a reverse calculation and try to find $A_c$ about the $IAOR$.

$A_c=\frac{v^2}{2r}$

$\space\space\space\space = \frac{4\omega^2r^2}{2r}$

$\space\space\space\space = 2\omega ^2 r$

And from equation $(1)$,

$A_c=2a_c$, which completely makes sense.

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  • $\begingroup$ I'm afraid but the point is not accelerating with 2w^2r but with w^r. This is justified because in the ground frame of reference the body is translating with a constant velocity. and if I jump to a frame rigidly attached to the centre of the disc I would observe the point on the disc to undergo a pure circular motion about the centre. But since I'm in an inertial frame( approx neglecting rotation effects of earth etc.) I would observe the same acceleration as with the ground that is w^2r. $\endgroup$ – user150098 Sep 15 '17 at 0:36
  • $\begingroup$ Although this might be true if the iaor is accelating upward with an acceleration w^2r. $\endgroup$ – user150098 Sep 15 '17 at 2:04
  • $\begingroup$ @gowreeshmago , Well IAOR indeed has an acceleration upwards. $\endgroup$ – Mitchell Sep 15 '17 at 4:55
  • $\begingroup$ Oh... I thought it was at rest wrt to ground. $\endgroup$ – user150098 Sep 15 '17 at 4:56
  • $\begingroup$ It is, but as you know centripetal acceleration cannot give speed to the particles. The POC for a purely rolling body will always be at rest, even though there is an acceleration upwards. $\endgroup$ – Mitchell Sep 15 '17 at 4:58
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The radius of curvature equals the distance to the instantaneous axis of rotation (IAOR). This is true due to the definition of the IAOR: At a given instant there is an axis perpendicular to the plane of motion such that any point of the body is in pure rotation, i.e. has circular trajectory about that axis. If a point is in circular motion about the IAOR, then the radius of curvature equals the distance to that axis as it is shown in the figure bellow:

enter image description here

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  • $\begingroup$ Thanks for your answer sir. But the radius of curvature depends upon the trajectory not the velocity( which I need more clarification about). The topmost point of the disc undergoes a trajectory recognized as cycloid and not a circle. The curvature of the circle is more than that of cycloid hence the radius of curvature is 4r not 2r. I've give n the proof in edit. $\endgroup$ – user150098 Sep 15 '17 at 0:30
  • $\begingroup$ @gowreeshmago Indeed the topmost point traces a cycloid but when we talk about IAOR we don't look to the path traced. We just take an infinitesimal time interval and in that interval every point on the disk traces an arc segment. $\endgroup$ – Diracology Sep 15 '17 at 1:08
  • $\begingroup$ If I assume what you are saying is correct here's a paradox. Suppose a disc is moving without slipping on a smooth horizontal surface. Whats the normal reaction by the ground. It's mg isn't it.The centre of mass translates w.r.t ground ( hence the radius of curvature is infinite) but according to you if the centre of mass rotates about iaor then the normal reaction by the ground would be less than mg according to the equation => Mg -N = mv^2/r. How would you explain this? $\endgroup$ – user150098 Sep 15 '17 at 1:16
  • $\begingroup$ And according to me the radius of curvature is infinite hence the rhs equals zero therefore N=Mg $\endgroup$ – user150098 Sep 15 '17 at 1:17
  • $\begingroup$ You're still looking at the trajectory when you shall look to an infinitesimal path. At a given instant there is an axis about the bottom most point and the whole disk rotates about that axis without translation!. $\endgroup$ – Diracology Sep 15 '17 at 1:45

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