2
$\begingroup$

I was solving a question where a disc was placed between two plates which were given velocities $V1$ and $V2$ and was asked to find radius of curvature at a point P on the disc. It was mentioned that there was no slipping.

Here A is a point on the lower plate and G represents the ground '-' represents with respect to

My teacher found velocity of centre of mass using the pure rolling condition and used instantaneous axis of rotation, considering it to be a line through the centre of the disc perpendicular to the plane. {He calculated velocity}

I have always had this thought that if any point on a rolling body has $v=0$ then it must be on the instantaneous axis of rotation. Is this true and if yes then would'nt there be another axis of rotation, one which passed throught the plane of disc touching the point of contacts with plates (where $v=0$ due to no slipping). But the answer with this approach is not matching with this approach.

Where am i wrong?

$\endgroup$
1
1
$\begingroup$

It depends on what $v$ is. When $v$ is the velocity of a particle as viewed by an inertial observer then yes where $v=0$ on the plane is where the axis of rotation goes through (in 3D it is a bit more complex because you can have translation parallel to the axis of rotation also).

Also note that the axis of rotation is instanteneous which means it can move around from one time frame to the other, and not in a smooth fashion.

But the no-slip condition means $v=0$ for relative velocity between contacting surfaces. In the example above both contact points have zero relative velocity and both have non-zero absolute velocity as seen by an observer.

So the $v_{\rm rel}=0$ on the contacts sets up the relative axis of rotation between the two contacting bodies. That is if you were moving along one of the plates you would observe the disk roll around the contact point with that plate, but not the other.

At any instant and from each observer, there is only one axis of rotation.

Move the observer from the bottom plate to the top plate and the axis of rotation changes. Or move the observer far away and find the axis of rotation at point P.

There is a theorem of kinematics that when two bodies are in contact their relative center of rotation has to be along the contact normal (even with slipping). And when three bodies are in contact the center of rotation of one body is always on the line connecting the relative centers of rotations with the other bodies.

In this case, the relative center of rotation are at (1) and (2) below, and so the axis of rotation of the disk must be on the line connecting them. You will find point (3) the axis of rotation does so.

fig1

$\endgroup$
1
$\begingroup$

I have always thought that if any point on a rolling body has $v=0$, then it must be the instantaneous axis of rotation.

This isn't correct.

Consider a cylinder rolling without slipping on a level surface amd moving with constant translational velocity $v$.

This means that the axis of rotation must be moving with translational velocity $v$. It also means that every point on the cylinder is also moving with the same translational velocity $v$.

However, we also need to consider the rotational motion. The condition of no slipping means that the surface of the cylinder is moving at speed $|v|$. Now, at the point of contact of the cylinder with the level surface, this rotational speed is directed in the opposite direction of the translational velocity. Hence they cancel, and we get an instantaneous vanishing velocity at the point of contact.

Moreover, a rigid body in 3 spatial dimensions can't have more than one instantaneous axes of rotation. (However, I can't resist adding that rigid bodies in 4 spatial dimensions can have two such axes ! Actually they are invariant rotational planes - the notion of an axis of rotation doesn't generalise to 4d - and generally speaking, they would be orthogonal to each other).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.