# Rotational and Translational Forces

1. Nov 27, 2013

### PBear

I am having a hard time proving this to myself:

Given some object with a known inertial moment and center of mass, with r = distance from COM, do forces of equal magnitude along some line L perpendicular to R all yield the same F_t (translational force) and F_r(rotational force), and how can I prove this? How can I prove that F_t and F_r are as I have shown?

The shape shouldn't matter, only that both forces are along line L and F_t and F_r are perpendicular with F_t pointing toward the COM. I get the feeling this isn't correct because with F being applied at the point r, the F_t would be 0. If it isn't correct, how can I find the relative forces of translation and rotation acting on some free body in the way that I described?

2. Nov 27, 2013

### Andrew Mason

Newton's second law applies. The problem is that in practice it is not possible to apply the same constant force to a single point on the perimeter as to the centre of mass (eg. equally to two points that are equidistant from the centre of mass on an axle through the centre of mass) because the force on the perimeter causes the body to experience angular acceleration (rotation) as soon as you apply it. This necessarily will change the direction of the applied force.

In order to apply Newton's laws, I think you will have to consider a force being applied to the same point in the reference frame of the centre of mass.

AM

3. Nov 28, 2013

### 256bits

Your diagram as drawn is incorrect though.

The point of application of F_r should at the same exact point of application of F_t, and not head to tail as shown. You would use the parallelogram law of vectors.

The application of F_t as drawn would be pure translation along its direction through the COM, and that is OK.
The force F_r would apply a translation of the COM in its direction of application. The vector sum of F_t and F_r yield the same translation as that from F.

And F_r would also apply a rotation. One would use the perpendicular of the direction of F_r to the COM. Thus, this would yield the same torque as F acting on its own.

PS. Moving a vectors position along its length does no changes.
Moving a vector perpendicular to its point of application - well then you have to add a torque in the opposite sense to counterbalance that which you have just produced.
What you could do with F is move it from its initial position, to a parallel location that will apply at the COM and then apply a couple(torque) to cause the rotation.

Last edited: Nov 28, 2013
4. Nov 28, 2013

### PBear

I should have been more clear regarding the force that was being applied. It is always to be applied distance r above the COM, in the direction shown. It is not applied to a fixed point on the object.

256Bits:
I was also trying to show the parallelogram law of vectors by applying F_t and F_r head to tail. You did clear up a lot regarding F_r also applying some translation. My question then is how to determine the amount of translation and rotation applied to a free body given such force. Is there a way to find the ratio of force applied to translation through the COM and pure rotation through r, given both I and r?

5. Nov 28, 2013

### Staff: Mentor

Your terms F_t and F_r are a bit confusing. Instead, think of a single force F, which creates both a translational acceleration of the center of mass and a rotational acceleration about the center of mass.

6. Nov 28, 2013

### PBear

That's what I'm attempting to do, but I would like to know how much rotational and translational acceleration. It should vary with r.

Edit: also the direction of the translation.

The reason I went down the path I did was because that was the answer I found on another forum: separating F into two forces, one through COM and one perpendicular, the one through com was pure translation and the other, F_r was pure rotation. From the answers I have gotten here, I now assume this is incorrect, though I know not why.

7. Nov 28, 2013

### Staff: Mentor

The translational acceleration is given by F = ma, and does not vary with r.

The rotational acceleration is given by Fr = Iα, which does depend on r.

8. Nov 28, 2013

### PBear

Assuming you mean F=ma_com, isn't the translational accelleration decreased because some of the force goes to rotation?

9. Nov 28, 2013

### Staff: Mentor

No, not at all. The entire force is involved in accelerating the center of mass. It doesn't matter where the force is applied, the same force gives the same translational acceleration. That's why I suggest you forget the idea of F_r and F_t and think in terms of just F.

10. Nov 28, 2013

### PBear

Ahh, I see. Thank you for clearing that up.