# Homework Help: Instantaneous Centre of Rotation

1. Jun 25, 2013

### andyrk

There is a rigid solid sphere rolling without slipping on a horizontal surface. When we are taking the CM as the axis of rotation, we see that the Total Kinetic Energy comes out be as:
KEtotal = KErotational + KEtranslational

Here:
KErotational = (1/2)ICMω2
where ICM is the Moment of Inertia of the rolling body about its CM and ω is the angular velocity of the particles constituting the rolling rigid solid sphere about the CM.

KEtranslational=(1/2)mV2CM

Here we use KErotational because every particle is rotating about the CM and we use KEtranslational because everything including the CM is moving forward linearly with velocity say VCM.

But when we take the point in contact with the ground as the instantaneous axis of rotation we only take KErotational.

Suppose the point of contact is called P. Then
KEtotal = KErotational = (1/2)IPω2
Where IP is the Moment of Inertia of the rolling body about the point of contact and ω is the angular velocity of the particles constituting the rigid rolling solid sphere about the point of contact.

I read that, about instantaneous centre of rotation all the particles rotate about it as if that is the centre of rotation and that point is pinned/fixed. So it behaves like a disc pinned at the centre which is rotating (only rotating since it can't translate as it is pinned) about its centre (CM). But how is this possible for rolling without slipping? I mean even the point of contact is instantaneously at rest how can it behave as pinned forever so that particles don't seem to translate at all? Also, how can there be simultaneously 2 axis of rotation possible? The 2 namely being: The CM and the point of contact.

2. Jun 25, 2013

I think the author is just trying to illustrate a point about rotational kinetic energy. Think about rotating about the center of the sphere. While it is rotating, all of the point of the sphere are constantly rotating about the center. To help you visualize the comparison, take the particles directly above the sphere's center. These points all have instantaneous velocity in the direction of motion. Now, take the frame where point P is the rotational center. If the sphere is translating, then instantaneously stops and now is thought to be 'rotating' about the contact point, well, all of the points of the circle are traveling in an analagous direction as those particles directly above the sphere were traveling when we rotated about the center of mass.

There arent two simultaneous axis of rotation. It is just putting the problem in a different perspective. Theres no way you could rotate about both points simultaneously. Once again, we are just putting ourselves in a different frame of reference and observing a problem from a different viewpoint. That is all.

3. Jun 25, 2013

### Staff: Mentor

The instantaneous axis of rotation is just that: The point that has zero velocity, so that for that instant you can treat the entire KE as pure rotational KE about that point. But that's good enough, since it still represents the total KE.

There's only one point about which the motion can be viewed as being in pure rotation. That's the instantaneous axis of rotation.

On the other hand, you can always view the motion as the sum of the translational motion of the center of mass and rotation about the center of mass. Either way will give you the total KE.

4. Jun 25, 2013

### andyrk

How do we know that if we take the KE about P (only rotational) we will get the total KE? How can we be sure that about point P we will get rotational KE which would equal the total KE as measured about the CM?

5. Jun 25, 2013

### andyrk

That's good. Only one point for pure rotation. But how many points for both translation and rotation? Like one of those points is the CM. Can there infinitely such points excluding the the point of contact? So I had a doubt then, that if we have multiple axis of rotations possible where about which rotation and translation can be viewed then suppose a sphere is lying on a floor which is smooth. We give it a torque by a force that we apply on its edge tangential to the surface and parallel to the floor. Depending on the axis we choose, we get different values of the torque being provided by the force. That means the body has different rotating capability at the same time of application of a single force. How is this possible? Lastly can there be ONLY one centre of pure rotation but many centres of both rotation and translation in rolling without slipping?

Last edited: Jun 26, 2013
6. Jun 26, 2013

### andyrk

"The combined effects of translation of the centre if mass and rotation about an axis through the centre of mass are equivalent to a pure rotation with the same angular speed about an axis passing through a point of zero velocity(Instantaneous Axis of Rotation)"
How? Why? Can we prove it?

7. Jun 26, 2013

### WannabeNewton

The proof can be found in many introductory mechanics books. See e.g. chapter 6 of Kleppner and Kolenkow "An Introduction to Mechanics" or just google it. Don't forget that torque only makes sense when measured with respect to a reference point; there is no absolute torque for a body. Things like angular velocity and angular acceleration on the other hand don't require a reference point in the above sense.

8. Jun 26, 2013

### andyrk

I searched for it but didn't find anything worthwhile. Can you post some link? Or can you explain it?

9. Jun 26, 2013

### WannabeNewton

Do you have something specific in mind? For example, say we have a rigid body translating and rotating with some angular velocity $\omega$ in a fixed plane. The kinetic energy in the lab frame is given by $T = \frac{1}{2}\int v^{2}dm$. Now transform to the center of mass coordinates so that $T = \frac{1}{2}\int (v' + V)^{2}dm = \frac{1}{2}\int v'^{2}dm + \frac{1}{2}MV^{2} + \int v'\cdot Vdm$ where $V$ is the center of mass velocity. Note that since every point on the rigid body has the same translational velocity as the center of mass, every point on the rigid body is simply rotating about the center of mass with angular velocity $\omega = \frac{v'}{r}$ as measured in the center of mass frame. Now $\int v'\cdot Vdm = V\cdot \int v'dm = V\cdot \frac{d}{dt}\int r'dm = 0$ because $\int r'dm = MR' = 0$ (the center of mass is at the origin in center of mass coordinates). Thus $T = \frac{1}{2}MV^{2} + \frac{1}{2}\omega^{2}\int r'^{2}dm = \frac{1}{2}MV^{2} + \frac{1}{2}I_{\text{CM}}\omega^{2}$ where $I_{CM}$ is the moment of inertia for an axis through the CM at a given instant.

On the other hand, let us say that at some instant of time the instantaneous center of rotation $O$ is some distance $R$ away from the center of mass of the rigid body. Note that in the frame centered on $O$, all points on the body instantaneously move on circles of respective radii about $O$ and since the center of mass must be on the body it too moves on a circle, with a velocity $V = \omega R$. The moment of inertia for an axis through $O$ will be given by $I_{O} = I_{CM} + MR^{2}$ (parallel axis theorem) so in this frame $T = \frac{1}{2}MR^{2}\omega^{2} + \frac{1}{2}(I_O - MR^{2})\omega^{2} = \frac{1}{2}MR^{2}\omega^{2} + \frac{1}{2}I_O\omega^{2} - \frac{1}{2}MR^{2}\omega^{2} = \frac{1}{2}I_{O}\omega^{2}$. So at this instant we can either calculate the kinetic energy by considering a pure rotation of the rigid body about the instantaneous center of rotation or calculate the kinetic energy by considering a pure rotation about the center of mass and the translation of the center of mass with respect to a lab frame.

10. Jun 26, 2013

### Staff: Mentor

About that instantaneous axis of rotation, you can view the body as being in pure rotation. So the rotational KE about that point is the total KE.

There's a theorem (given in most Classical Mechanics textbooks) that the total KE of a rigid body is given by the sum of the translational KE of the center of mass plus the rotational KE about the center of mass. (This is what WannabeNewton is describing.)

11. Jun 26, 2013

### andyrk

How did

$\frac{1}{2}(I_O - MR^{2})\omega^{2}$​
come up? Actually I din't get the whole equation where this is used! :/
Thanks though! :)

12. Jun 26, 2013

### andyrk

Is that theorem for highschool level? Or undergraduate studies?

13. Jun 26, 2013

### D H

Staff Emeritus
Classical mechanics is the class physics majors take after completing the standard calculus-based introductory physics classes (two to four classes, depending on the college). A typical classical mechanics class assumes knowledge of differential equations and linear algebra. That's the mathematics one studies after taking the standard calculus mathematics classes.

In short, it's not high school level.

14. Jun 26, 2013

### WannabeNewton

Substitute for $I_{CM}$ using the expression obtained from the parallel axis theorem.

15. Jun 27, 2013

### andyrk

How do you know that the expression of T that you got only includes rotational KE and noth rotational and tranlsational both?

Why was there a need to write:
$T = \frac{1}{2}MR^{2}\omega^{2} + \frac{1}{2}(I_O - MR^{2})\omega^{2}$​

Didn't get it..

16. Jun 27, 2013

### WannabeNewton

The expression for KE based on quantities measured with respect to the instantaneous center of rotation is simply $T = \frac{1}{2}I_{O}\omega^{2}$. This is pure rotation-there is no translation term as you can easily see.

17. Jun 27, 2013

### andyrk

If you were measuring motion about the instantaneous point of contact that is point O then why didn't you simply write:
$T = \frac{1}{2}I_{O}\omega^{2}$​
$T = \frac{1}{2}MR^{2}\omega^{2} + \frac{1}{2}(I_O - MR^{2})\omega^{2}$​
this first and then equating it to $\frac{1}{2}I_{O}\omega^{2}$?

18. Jun 27, 2013

### WannabeNewton

Because I was proving how you derive the result based on the expression written down in terms of quantities measured with respect to the CM. You can't simply write something down if you don't prove that it's true in the first place.

19. Jun 27, 2013

### andyrk

Co-ordinates of CM are defined as:
$\int r'dm$/M

This has to be 0 for the co-ordinates of CM to be 0. How does this become 0?

Last edited: Jun 27, 2013
20. Jun 27, 2013

### WannabeNewton

$R'$ represents the position vector from the origin of the center of mass frame to the center of mass, which is just the origin itself so it vanishes.

21. Jun 27, 2013

### andyrk

How did you straight away jump to this result is what is bugging me. :/

22. Jun 27, 2013

### WannabeNewton

We have $T = \frac{1}{2}MV^{2} + \frac{1}{2}I_{CM}\omega^{2}$. With respect to the instantaneous center of rotation $O$, we can write $V = \omega R$ where $R$ is the distance from $O$ to the CM. We can also use the parallel axis theorem to write $I_{O} = I_{CM} + MR^{2}$ hence $T = \frac{1}{2}MR^{2}\omega^{2} + \frac{1}{2}(I_O - MR^{2})\omega^{2} = \frac{1}{2}I_{O}\omega^{2}$.

23. Jun 27, 2013

### WannabeNewton

Note that you can also show $T = \frac{1}{2}I_{O}\omega^{2}$ directly since $T = \frac{1}{2}\int v^{2}dm = \frac{1}{2}\int r^{2}\omega^{2}dm = \frac{1}{2}\omega^{2}\int r^{2}dm = \frac{1}{2} I_{O}\omega^{2}$ when measured about the instantaneous center of rotation $O$. I was just showing how you can also get it out of the KE written in terms of CM quantities (after first proving the CM result) because it seemed like that's what you wanted.

24. Jun 27, 2013

### andyrk

What would be the limits of $\int r'dm/M$? And when $\int r'dm$ is expanded in the form of a sigma(Ʃ)
then how does it become $MR'$?

25. Jun 27, 2013