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.
-
$\begingroup$ 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? $\endgroup$– Maury MarkowitzApr 24, 2018 at 14:00
-
$\begingroup$ Related: Two axes for rotational motion and links therein. $\endgroup$– Qmechanic ♦Apr 24, 2018 at 14:27
-
4$\begingroup$ Possible duplicate of Two axes for rotational motion $\endgroup$– sammy gerbilApr 25, 2018 at 16:34
-
$\begingroup$ no the above links are not the answers to my question,they are questions of similar words but are of totally different conditions. $\endgroup$– sachinApr 26, 2018 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.
-
$\begingroup$ 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. $\endgroup$– sachinApr 24, 2018 at 18:06
-
$\begingroup$ 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. $\endgroup$ Apr 24, 2018 at 23:10
-
$\begingroup$ @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 ? $\endgroup$ Jul 2, 2020 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.
-
$\begingroup$ 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. $\endgroup$– sachinApr 26, 2018 at 10:45
-
$\begingroup$ @sachin - I along with others don't understand your concern. You are creating an issue where no issue exists. $\endgroup$ Apr 26, 2018 at 11:31
-
$\begingroup$ 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. $\endgroup$– sachinApr 26, 2018 at 11:48
-
$\begingroup$ @ 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 quora.com/Can-you-arbitrarily-place-the-axis-of-rotation. $\endgroup$– sachinApr 27, 2018 at 18:43
-
1$\begingroup$ So long as the rod wasn't rotating to begin with, @Shashaank, you are correct. $\endgroup$ Jul 2, 2020 at 13:14
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.