Let's say a body is undergoing both rotational and translational motion.

I know that ICR of the body as a whole will be the point about which the body is doing pure rotation, so basically will be the point with zero velocity, and that it will lie where perpendiculars to the velocities of each point intersect. I heard about a concept of ICR of each point, defined as "The point about which the given point rotates". Can you please explain this concept, of individual ICRs? Also, how can we calculate its location, and is it same as the centre of curvature for that point?

Please take a look at this - ICR == Centre of Curvature

As you can see, the ICR of the yellow point lies on the yellow broken line, and its distance is given by 4R and not 2R. Similarly of the green one is double that of its distance from the common ICR. The point where they both intersect is the ICR of the body as a whole, but then the individual points have different ICRs too. Please explain.

  • $\begingroup$ A body doesn't rotate about a point but about an axis. $\endgroup$ – Gert Oct 16 '15 at 16:46
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    $\begingroup$ Yes I'm sorry. I meant that ICR is the point on the plane through which the perpendicular axis of rotation passes. (Is that correct?) $\endgroup$ – Shodai Oct 16 '15 at 16:47
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    $\begingroup$ Alright, it's taken me a couple of tries but I see what you're doing wrong. Basically, the following should be true: $R=v^2/a_c=(\omega R)^2/(\omega^2 R)=R$, but you're assuming that $a_c=\omega R$ no matter what point you calculate it about. $a_c=\omega^2 R$ around the center of the disk, but around the point you've specified, $a_c=\omega^2 2R$. Also, for point B, $a_c=\omega^2 (\sqrt{2} R$). In any case, you should not expect those two to be the same, since A and B are in different locations anyway. $\endgroup$ – levitopher Oct 18 '15 at 3:35
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    $\begingroup$ Since the radius of curvature is the distance from the point in question to the point of rotation (that's what $a_c=v^2/R$ means), you can simply use geometry to find these answers. From A to the point is $2R$, and from B to that point is $\sqrt{2}R$. Both of these correctly describe the distance from A and B to the ICR, so everything works fine (as we should expect!) $\endgroup$ – levitopher Oct 18 '15 at 3:37
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    $\begingroup$ Yes, that's correct. In general the angular speed (and acceleration) will depend on which point you choose to calculate it about. If you pick the ICR, it will be constant and equal to $v^2/R$. If you pick a different point, it will be non-constant in time. $\endgroup$ – levitopher Oct 19 '15 at 2:37

Well, you've answered yourself a little bit. The instantaneous center of rotation is the point about which "the whole body" is performing pure rotational motion, so the ICR of each individual point of that body will be the same as the ICR for the entire body. To find this ICR, I think this image from wikipedia says it all:

"Pole-object-A1-A2" by VanBuren - Own work. Licensed under CC BY-SA 3.0 via Commons.

You take two points in the body, and find their velocity vectors. Bisect each velocity vector, and the intersection point of the bisections is the ICR. Therefore, to find the ICR, you need at least two points in the body, but once you've found it, you can speak of "the ICR of each point in the body".

The center of curvature is the point about which a point is moving in a circle. Since we find the ICR by using perpendicular vectors to the velocity vectors, the velocity vectors are tangents to a circle of radius equal to the distance to the ICR. Therefore, the ICR is the instantaneous center of curvature.

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    $\begingroup$ Hi, okay my second question has been cleared, thanks. But I'm still a bit confused about the first part. Our teacher told us that for finding the ICR of the entire body, we must draw normals to atleast 2 velocity vectors. The point where they meet will be the ICR. But for the ICR of individual points, he said that it is a different concept, and that we must draw a normal to the velocity vector, and the distance on that normal will be given by v^2/a (Radius of curvature). And it does come out to be different. So can a point have 2 ICRs? $\endgroup$ – Shodai Oct 16 '15 at 16:52
  • $\begingroup$ I believe the radius of curvature is the same as the IRC (and you instructor is incorrect). The IRC is the point at which every other point in the body is moving perpendicularly with respect to - this is also the same condition for circular motion. The acceleration for circular motion is $a_c=v^2/R$, where $v$ is the tangential speed. Since both definitions require circular motion, the IRC and $v^2/a_x$ should give you the same result. Can you provide a counterexample? $\endgroup$ – levitopher Oct 18 '15 at 3:13
  • $\begingroup$ Further clarification, based on your example (once corrected, see what I said up there). The IRC is a point; it will have a different distance between it and each point in the body. The radius of curvature is a distance; it will be the distance between the point in question and the IRC. $\endgroup$ – levitopher Oct 18 '15 at 3:25
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    $\begingroup$ Thank you sir this was helpful to me as well. But like the op many sites on the internet also give the radius of curvature at the top point to be 4R for example. Are all these websites wrong sir? $\endgroup$ – JustJohan Jan 26 at 9:13
  • $\begingroup$ @JustJohan: In this case, the question is technically well-posed for the instant in question, but not at any other instant of time. At the very moment shown in the picture, the radius of curvature of the trajectory of point A if $4R$. However, the point in question is actually moving in a cycloid: en.wikipedia.org/wiki/Cycloid, not a circle. It's very confusing to refer to "the center of curvature of a cycloid", because the answer depends on what part of the cycloid you are referring to. $\endgroup$ – levitopher Jan 26 at 15:49

In the 2D case each point (let's call it A) on the rigid body has two velocity components $(v_x,v_y)$ and the body itself rotates by $\omega$. The location of the ICR relative to A is $$icr = \begin{pmatrix} -\frac{v_y}{\omega} & \frac{v_x}{\omega} \end{pmatrix}$$

To prove this check that $\vec{v}_{icr} = \vec{v}_A + \vec{\omega} \times \vec{r} = 0$ or with planar vectors $$\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} v_x \\ v_y \\ 0 \end{pmatrix} + \begin{pmatrix}0\\0\\ \omega \end{pmatrix} \times \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} $$ $$ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} v_x - \omega \,y \\ v_y + \omega \,x \\ 0 \end{pmatrix} $$

$$ \begin{matrix} x=-\frac{v_y}{\omega} \\ y=\frac{v_x}{\omega} \end{matrix} $$

Also, since the body is rotating about the ICR the point A prescribes a circle (instantaneously) and thus yes the ICR is the center of curvature of all points on the rigid body.

In the 3D case a rigid body will rotate about an axis and it might have a parallel translation to that axis as well. Given a point A (on the body) with velocity $\vec{v}_A = (v_x,v_y,v_z)$ and rotational velocity of the body $\vec{\omega} = (\omega_x,\omega_y,\omega_z)$ the location if the ICR is given by

$$\vec{r}_{icr} = \frac{\vec{\omega} \times \vec{v}_A}{\|\vec{\omega}\|^2} $$

In addition, the direction of rotation is $\vec{e} = \frac{\vec{\omega}}{\|\vec{\omega}\|}$ and the vector of any parallel translational motion is $$\vec{v}_\parallel = \left( \frac{\vec{\omega} \cdot \vec{v}_A}{\|\vec{\omega}\|^2}\right) \vec{\omega}$$ The scaling factor in the parenthesis above is called the screw pitch because the motion of rotation with parallel translation is a screw motion.

To prove this, look at the well known transformation law $\vec{v}_{icr} = \vec{v}_A + \vec{\omega} \times \vec{r}_{icr} =0 $ and cross each side with $\vec{\omega}$

$$ \begin{array}{rcl}0&=&\vec{\omega}\times\vec{v}_{A}+\vec{\omega}\times\left(\vec{\omega}\times\vec{r}_{icr}\right)\\0&=&\vec{\omega}\times\vec{v}_{A}+\vec{\omega}\left(\vec{\omega}\cdot\vec{r}_{icr}\right)-\vec{r}_{icr}\left(\vec{\omega}\cdot\vec{\omega}\right)\\\vec{r}_{icr}\left(\vec{\omega}\cdot\vec{\omega}\right)&=&\vec{\omega}\times\vec{v}_{A}\\\vec{r}_{icr}&=&\frac{\vec{\omega}\times\vec{v}_{A}}{\vec{\omega}\cdot\vec{\omega}}=\frac{\vec{\omega} \times \vec{v}_A}{\|\vec{\omega}\|^2} \end{array} $$

In step 3 above it is assumed that $\vec{\omega}\cdot \vec{r}_{icr} = 0$ which turns out to be true given that $\vec{r}_{icr}$ is perpendicular to the rotation axis.

  • $\begingroup$ Hi, I didn't understand how you used the determinants (I don't know what they are), but I can understand somethings. If I'm only talking about 2-D Motion, then can a body have 2 ICRs? I am editing my question and adding an image to illustrate my doubt, please see that, thanks. $\endgroup$ – Shodai Oct 17 '15 at 2:34
  • $\begingroup$ No each rigid body only has one unique center of rotation. I am not using determinants, just vectors. The $\times$ is the vector cross product and $\cdot$ the vector dot product. The notation $| \vec{a} |$ means the magnitude of $a = \sqrt{a_x^2+a_y^2+a_z^2}$ $\endgroup$ – John Alexiou Oct 17 '15 at 5:31
  • $\begingroup$ Sorry I just figured that. But please see my figure. I have calculated Centre of Curvature and it's coming to be different from the ICR from the body. I am not saying that a rigid body has 2 ICRs, but that a point in the body has 2 ICRs - One is the ICR of the body, and other is the Centre of Curvature of that point. $\endgroup$ – Shodai Oct 17 '15 at 8:03
  • $\begingroup$ ja72: $$\vec r_{ICR} := \frac{\vec \omega \times \vec v_A}{\| \vec \omega \|^2}.$$ -- I wonder whether/how this fits with $$\vec v_A = \vec \omega \times (\vec r_A - \vec r_{ICR}) + \vec v_{ICR} $$ ... $\endgroup$ – user12262 Oct 17 '15 at 13:04
  • $\begingroup$ $\vec{v}_{ICR}$ should be zero by definition. If you plug in $\vec{r}_{ICR}$ in the transformation above you will get the known $\vec{v}_A$. Note that use have to use the vector triple product identity $$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a}\cdot \vec{c}) - \vec{c} ( \vec{a}\cdot \vec{b})$$ I show it in the proof section of the linked answer physics.stackexchange.com/a/95542/392 $\endgroup$ – John Alexiou Oct 17 '15 at 20:46

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