# Rotational Freefall and Designation of Axis

1. May 26, 2004

### WhackyWookie

Well, I'm not even sure if freefall is the right word, but my first quandary is as followed:

If two objects fall from the same height, and one object will begin to rotate after it has been release (ie. a yoyo), while the other will not rotate, then the rotating object will accelerate slower and hit the ground after the non-rotated one because of conservation of energy (PE = translational KE + Rotational KE).

But I can't seem to justify the slower acceleration of the rotating object in the force point of view since, based on my understanding, the torque of the rotating object should not affect the net force on the CM, therefore, the net forces on the CMs of both object should be the same and the translational acceleration should then also be equal.

I know there is something wrong with my logic, but I can't find where. Any help would be much appreciated.

My second question has to do with a quote in my physics textbook: "we see that if a body is to remain at rest, the net torque applied to it (calculated about any axis) must be zero .... the choice of axis is arbitrary. If a body is at rest, then net torque = zero about any axis what so ever."

What I do not understand is why can the axis be arbitrary is definition of torque is force times distance to the axis? For example, if a large rigid stick has two equally strong torque of opposite direction acting on the two ends and one force that cancel out the translational effect of the torque acting on the CM, then the stick will remain at rest if and only if the axis of rotation is the CM, otherwise, the torque will be unbalanced and it will spin.

Any clarification is again, much appreciated. Thanks

2. May 26, 2004

### arildno

1. Careful!
A yo-yo does not experience free-fall; the cord imparts an upwards force on the yo-yo acting tangentially,in the stretched cord's direction on the axis; hence the spinning, and the slower acceleration.

2.Here, you have not accounted for the moments of ALL forces around an arbitrary axis properly.
The argument is as follows:
Suppose we have balance of 3 forces, and balance of the three moments about a given axis, i.e:
$$\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=\vec{0},\vec{r}_{1}\times\vec{F}_{1}+\vec{r}_{2}\times\vec{F}_{2}+\vec{r}_{3}\times\vec{F}_{3}=\vec{0}$$

Let us now consider the moments of the same forces about a differently situated axis.
The new position vectors to the points where the forces acts, are therefore:
$$\vec{r}_{N,1}=\vec{r}_{1}-\vec{x}_{A},\vec{r}_{N,2}=\vec{r}_{2}-\vec{x}_{A},\vec{r}_{N,3}=\vec{r}_{3}-\vec{x}_{A}$$

(Here, $$\vec{x}_{A}$$ is the position vector of the "new" axis in the "old" coordinate system)

Computing the moment about the new axis yields:
$$\vec{r}_{N,1}\times\vec{F}_{1}+\vec{r}_{N,2}\times\vec{F}_{2}+\vec{r}_{N,3}\times\vec{F}_{3}=$$
$$\vec{r}_{1}\times\vec{F}_{1}+\vec{r}_{2}\times\vec{F}_{2}+\vec{r}_{3}\times\vec{F}_{3}-\vec{x}_{A}\times(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3})=\vec{0}$$

Last edited: May 26, 2004
3. May 26, 2004

### turin

I will let arildno's response to your first question speak for myself as well.

This is not quite the definition of [applied] torque. The definition should involve a cross product at least.

You are probably thinking of torque as some kind of extension or generalization of force. You should instead think of torque as the rotational analog of force. Both "force" and "torque" are subsets of the generalized/canonical force. Anyway, the point is that applying torque to a body already implies that you are not applying a translational/linear force.

There is a subtle difference in an applied force that induces a net torque about the CM vs. an applied torque to begin with which implies no applied linear force.

Here is a word document with some illustrations that may help:

#### Attached Files:

• ###### applied_vs_induced_torque.doc
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4. May 28, 2004

### WhackyWookie

I understand arildno's explanation for the 2nd question. can't believe i overlooked that.

But the rest I'm still somewhat vague about.

can you guys elaborate a bit more in context to this question:
If one was to impart a force on the end of a pen in space at 90 degree, would the pen move forward and rotate or just rotate in the same position?

5. May 28, 2004

### Janitor

It would both move forward (i.e. its CM would translate) and it would rotate.

If you did not want the CM to move, you could get pure rotation by using two fingers, one at each end of the pen, pushing equally hard in opposite directions. (I am assuming an ideal pen with appropriate symmetries.)