I m not concerned with rotation of a body with two simultaneous axis but concerned with how we choose the axis,while going through pure rolling I have observed that there are two axis of rotation one is passing through the center of mass and the other is through the point in contact with the ground,my concern is how can there be any axis of rotation through the point of contact where as m very well finding the body does not rotate in that axis of rotation that is it very well rotates only through the center of mass.
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Your question is not clear. What is "pure rolling"? And what is this second axis at the point of contact you are referring to. Perhaps you could draw it? – Maury Markowitz Apr 24 '18 at 14:00
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Related: Two axes for rotational motion and links therein. – Qmechanic Apr 24 '18 at 14:27
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4Possible duplicate of Two axes for rotational motion – sammy gerbil Apr 25 '18 at 16:34
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no the above links are not the answers to my question,they are questions of similar words but are of totally different conditions. – sachin Apr 26 '18 at 10:43
3 Answers
A body will only have one instantaneous axis of rotation (Chasle's Theorem).
In your example, the center of mass translates, and thus it is not a center of rotation. The only center of rotation is the point that does not move (the contact point).
To find the location of the center of rotation relative to some point A where you know the velocity vector $\boldsymbol{v}_A$ for a body with rotational velocity vector $\boldsymbol{\omega}$, calculate the following
$$ \boldsymbol{r}_{\rm rot} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_A }{ \| \boldsymbol{\omega} \|^2} $$
Here $\times$ is the vector cross product.
So consider the motion of the center $v$ to the right, and the no-slip condition $v + \dot{\theta} R =0$. In vector form the motion of the center is
$$ \begin{aligned} \boldsymbol{v}_A & = \pmatrix{v \\ 0 \\0}& \boldsymbol{\omega} & = \pmatrix{0 \\ 0 \\ -\frac{v}{R}} \end{aligned} $$
The center of rotation is thus
$$ \boldsymbol{r}_{\rm rot} = \frac{ \pmatrix{0 \\ 0 \\ -\frac{v}{R}} \times \pmatrix{v \\ 0 \\0} }{ \| \pmatrix{0 \\ 0 \\ -\frac{v}{R}} \|^2 } = \frac{ \pmatrix{0 \\ -\frac{v^2}{R}\\0}}{ \left( \frac{v}{R} \right)^2} = \pmatrix{0 \\ -R \\ 0}$$
which is in fact the contact point.

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if the center of rotation is the contact point,why the body does not move along it that is why it does not enter inside the earth to make a circle and move with a radius of rotation equal to the diameter of the sphere,I hope m clear that i mean the body moves only on the earth and not getting hinged at the contact point. – sachin Apr 24 '18 at 18:06
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By definition the rotation center is the location where the body doesn't have any linear velocity. But the center itself can move with time. That is why it is called the instantaneous center of rotation. If a wheel is slipping then the rotation center is somewhere on the vertical line (contact normal) such that the velocity doesn't penetrate into the ground. The no slip conditions forces the center to meet with the ground. – John Alexiou Apr 24 '18 at 23:10
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@JohnAlexiou I have a small doubt here. If you take your body coordinate system at the centre of mass (or any other point too), how would you calculate the angular velocity of the rigid body. To calculate the angular velocity of the rigid body you need to know the instantaneous axis of rotation and the velocity of a point of the body. Is there any other general method of calculating the angular velocity of the rigid body without knowing the instantaneous axis of rotation ? Could you please help me in this regard ? – Shashaank Jul 02 '20 at 13:14
... it very well rotates only through the center of mass.
This is not the case. In some applications, it's most convenient to describe the motion of a rigid body as a combination of translation by the center of mass and rotation about an axis passing through the center of mass. You appear to have fallen into the trap of thinking that this is the only way to describe the motion of the rigid body. In other applications, it's even more convenient to describe the motion as a combination of translation of some other central point and rotation about an axis passing through that point.
A rigid body can be viewed as having an infinite (uncountably infinite) number of axes of rotation. Suppose you know the velocity of some central point $c$ of the object and the object's angular velocity. The velocity of some other point of the rigid body $p$ is $\boldsymbol v_p = \boldsymbol v_c + \boldsymbol \omega \times (\boldsymbol r_p - \boldsymbol r_c)$. There's nothing special about the center of mass in this construction. You can pick any arbitrary point on, inside, or even outside the body as the central point.
Another way to look at it: Angular velocity is a free vector. It's the same for every point inside or on the rigid body.

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my concern is still unclear,can the sphere be taken to be hinged at the point of contact and revolving around it when we say the axis of rotation is passing through the contact point. – sachin Apr 26 '18 at 10:45
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@sachin - I along with others don't understand your concern. You are creating an issue where no issue exists. – David Hammen Apr 26 '18 at 11:31
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my issue is simple,I understand that an axis of rotation is the line about which all the points of a body move in a circle what stands well when m taking the center of mass as the point of rotation but makes absurd sense when m considering the contact point as the point of rotation,hope m clear. – sachin Apr 26 '18 at 11:48
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@ David Hammen "A rigid body can be viewed as having an infinite (uncountably infinite) number of axes of rotation" basically means i can choose any arbitrary axis of rotation,so a wheel is rotating and i can say its axis of rotation may be anywhere not necessarily passing through the center,the concept is violated in this discussion https://www.quora.com/Can-you-arbitrarily-place-the-axis-of-rotation. – sachin Apr 27 '18 at 18:43
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@DavidHammen. I have a small doubt here. You write "If the angular velocity of the body is know.." If you take your body coordinate system at the centre of mass (or any other point too), how would you calculate the angular velocity of the rigid body. To calculate the angular velocity of the rigid body you need to know the instantaneous axis of rotation and the velocity of a point of the body. Is there any other general method of calculating the angular velocity of the rigid body without knowing the instantaneous axis of rotation ? Could you please help me in this regard ? – Shashaank Jun 28 '20 at 17:06
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Can you let me this point because I won't be able to ask a new question citing just this specific point..... – Shashaank Jul 01 '20 at 19:26
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@Shashaank - You do not need to know the instantaneous axis of rotation. No matter where one places a rate gyro anywhere on a rotating rigid body, it will register the same rotation with respect to inertial. It's much better to think of an infinite number of instantaneous axes (note the use of the plural) of rotation as a bundle of parallel lines. – David Hammen Jul 02 '20 at 02:56
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@sachin - This is a very late response. The above comment also applies to your comment. – David Hammen Jul 02 '20 at 02:57
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@DavidHammen ok thanks for your comment. I had thought what you are saying but then the only doubt I was having was how will one calculate the omega for the rigid body in the first place... Is it that- place your axis anywhere inside the body and just figure the amount a point in the body rotates about this axis ($d\theta/dt$) and then you are done. Is that correct... Is that the best way to find omega in a general problem and then figure out other quantities like kinetic energy. Is is what you are also trying to say in your above comment, right. – Shashaank Jul 02 '20 at 04:53
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@DavidHammen like for example. You have a rod in space. You apply a force at its COM. It's a rigid body diaplacement. So you can it be described by a translation and a rotation about any point ( Euler's theorem)... So we see the rod only translates. But if you apply Euler's theorem then you need to figure out what omega is. My guess (according to what you said).- calculate omega about any point, inside, about COM or any other point, it will always turn out be 0. Is that correct. Have I got the total concept correctly or is anything still wrong – Shashaank Jul 02 '20 at 05:10
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@DavidHammen what I have understood is correct, right or anything wrong in that.... – Shashaank Jul 02 '20 at 13:10
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1So long as the rod wasn't rotating to begin with, @Shashaank, you are correct. – David Hammen Jul 02 '20 at 13:14
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@DavidHammen thanks a lot for you help... Finally I have got it... Just a confirmation - will omega be the same even if the axis is outside the body or not. Like $omega = d\theta/dt$, if you measure $d\theta$ about any axis through any point inside the body you get the omega which will be same about any point inside. Will it be same for any point outside too. I guess yes because instantaneous axis of rotation can be outside too. Is this right too finally – Shashaank Jul 02 '20 at 13:22
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@DavidHammen I think it's correct.. Maybe you can just confirm for the best – Shashaank Jul 02 '20 at 13:24
It isn't so. You can't have two axes at once. You can combine rotation along two axes to give a effective rotation along a single axis and the other way round.