# Angular velocity of rigid body

1. Jan 11, 2014

### arvindsharma

Dear All,

i was reading a book in which it is written that if a rigid body have both rotational(with angular velocity 'w' about its center of mass)and translation(with velocity 'v' of center of mass) motion then the whole motion can be treated as an equivalent motion of pure rotation about instantaneous center of rotation with same angular velocity 'w'.my doubt is that how can angular velocity be same as we are changing our reference point and angular velocity is a function of reference point?please explain this and if possible give me the mathematical proof of this.i am trying to prove it since last two days but failed every time.

Thanks
Arvind

2. Jan 11, 2014

### Simon Bridge

The reference point for w is changing - that's why w stays the same.

3. Jan 11, 2014

### BruceW

so, we simply use a frame of reference where the centre of mass is (instantaneously) stationary? Is this what it means? for example, if the centre of mass of the rigid object is moving at some speed $v$ away from us, then if we run towards it at speed $v$, then we no longer see its translational motion, we only see its rotational motion. I'd guess that's what it means..

4. Jan 11, 2014

### Tanya Sharma

Suppose a rigid body is rotating about a vertical axis passing through point O in the attached figure .Let us try and show that the angular velocity is same for two reference points O and P.

Consider two points on the rigid body P and A as shown in the figure. The velocity of point A is

vA = ωO x r = ωO x (a + r') = vP + ωO x r' (1)

But we know that vA = vA w.r.t P + vP i.e ωP x r' + vP where ωP is the angular velocity of the body with respect to P .

or, vA = vP + ωP x r' (2)

Comparing eq 1 and 2 we can deduce that ωO = ωP .That is, the angular velocity about point O and P is the same.

You can see that the angular velocity ω of a rigid body about any point is same.

Hope that helps

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Last edited: Jan 11, 2014
5. Jan 11, 2014

### Staff: Mentor

No. The instantaneous axis of rotation is a point in the body that is at rest (at that instant) and thus the entire motion of the body can be considered (at that instant) as in pure rotation about that point.

6. Jan 11, 2014

### BruceW

well, my suggestion would be sufficient to make the entire motion of the body be considered as in pure rotation about that point, right? So do you mean that my suggestion is overly specific? So maybe as a more general condition, all we need to do is change frame of reference such that a given point in the body is instantaneously at rest, and that gives us (instantaneously) pure rotation.

edit: i.e. the point we choose to be instantaneously at rest could be any point in the rigid object.

7. Jan 11, 2014

### Staff: Mentor

It's not about changing frames, but picking the point about which the body is instantaneously in pure rotation about. (In the same frame.)

8. Jan 11, 2014

### BruceW

uh... I'm answering the OP's question, which was If there is a rigid object with both translational and rotational motion, then this can be treated as purely rotational motion. This is not great wording, but I'm guessing it means that we can always change frame of reference by choosing a frame of reference with different velocity, and such that all points on the rigid body are in instantaneous pure rotation about a single axis. Although maybe I misunderstood the OP'er, who hasn't replied yet unfortunately.

9. Jan 11, 2014

### Staff: Mentor

Yes, purely rotational motion about the instantaneous center of rotation. Look it up!

10. Jan 11, 2014

### BruceW

I assumed the OP's problem was about a general 3d object, therefore (generally) there is no instantaneous center of rotation. But I guess it is possible the OP'er was talking about planar motion. hard to tell, since he hasn't replied.

11. Jan 11, 2014

### Simon Bridge

You can use any inertial frame and the physics is unchanged.

Let an object translate and at the same time rotate about it's center of mass.
Put a dot at point D, on the object, (a distance R from it's center of mass, say).

If the origin is point O and the center of mass is at point C,
then I can define some vectors: $$\vec r_d=\overrightarrow{OD},\; \vec r_c=\overrightarrow{OC},\; \vec r_{cd}=\overrightarrow{CD}$$ Given this, then the statement:
... is trying to say[*] the following:$$\vec r_d=\vec r_c + \vec r_{cd}$$ The relationship should be clear from the definitions.

From there the algebra should show you the rest as long as you are careful when you define the angular velocities. If you are having trouble, you should show your working.

You can demonstrate it by tossing something in the air while giving it a bit of a spin - a brick say.
Watch carefully - follow the center of mass with your eye ... it should be easy to convince yourself that it follows a parabola. More carefully, can you see that the brick keeps rotating about the center of mass at the same rate that you gave it initially? It helps if you just toss it a bit and catch it again so the top of the arc is in front of your face.

Note: In general the rotation-axis (through the center of mass) can have any orientation wrt the direction of motion.

------------------------

[*] as written, the statement is flat wrong - but I am assuming it just got garbled.
The vector equation is what they should have said: the motion can be treated as a combination of pure rotation and pure translation.

Last edited: Jan 12, 2014
12. Jan 12, 2014

### arvindsharma

thank you all.now my doubt is clear.

13. Jan 12, 2014

### Simon Bridge

Excellent, well done.

14. Jan 13, 2014

### Staff: Mentor

Since the OP explicitly mentioned the instantaneous center of rotation I assumed that was the case.

I think the real issue was not the existence of an instantaneous center of rotation (or changing to a different frame), but why the rotational speed is the same taken about any point. That was nicely handled by Tanya Sharma.

15. Jan 13, 2014

### BruceW

ah whoops, yes he does say that.