# Translational Acceleration with No Rotation

1. May 14, 2015

### Zalajbeg

Hello Everyone,

Let us think about a rod whose length is L and end points are A and B. The road is parallel to the ground therefore the weight of the rod is perpendicular to the line between A and B.

The weight of the rod is F and two forces with the magnitude of F are exerted on both ends and the direction of these forces are opposite to the direction of weight.

The vertical force balance: 2F-F = m.a, therefore there is an acceleration towards the upper side.

However the torque balance makes me confuse. Which point do I need to use? If I use the mid-point the total torque is zero. However if I use one of the end points the magnitude of total torque is FL/2. Then what does it mean? Does it have an angular acceleration or not?

I thought about that maybe it means there is no angular acceleration around the mid-point axis and there is an acceleration around the end point axis. However if there is no initial angular velocity and no angular acceleration around the mid-point axis the relative positions of the points on the rod seem constant to me. Therefore it sounds ridiculous to me to say there is an acceleration around end points if there is no acceleration around mid-point.

Could someone please explain me what I am missing?

2. May 14, 2015

### PeroK

You're missing two things: that the weight also exerts a torque about points other than the centre of mass; and, that a body moving in a straight line has non-zero angular momentum about any point not on that line.

3. May 14, 2015

### Quotes

I think its due to difference in reference system. If the rod isn't fixed to any point the rotational would take about the centre of mass we will take CM to be the origin . And in this case net torque is zero as the torque due to the forces at the ends are equal and opposite..
But if the rod is pivoted to a point at the end say A (we will now consider this as the origin) force at this point will give a null torque. There will two other torqe one due to weight say T1 and other due to force at B say T2... They both would be in opposite directions but T2>T1 thus there would be a net torque giving the rod a net angular acceleration.....
Hope this helps

4. May 14, 2015

### Zalajbeg

I think I am not missing the first thing. As I say the torque around the point A is FL/2 I take the torque effect of the weight into consideration (FL-FL/2=FL/2).

1) A body moving in a straight line.. What is this line? Is it the line which center of mass moves one it?

2) A body moving in a straight line.. Did you really mean move or acceleration? Also you spoke about the angular momentum actually I am not able to relate it to angular acceleration. Can you please clarify a bit more?

Thanks for your effort. I will try to remember my knowledge about using different coordinate systems.

5. May 14, 2015

### Quotes

Nd for the answer of your first question..... I have read in book that- "translational motion of an object can be understood as the motion of the centre of mass of the object"... So yes this would imply on straight line motion also

6. May 14, 2015

### Quotes

Nd for the answer of your first question..... I have read in book that- "translational motion of an object can be understood as the motion of the centre of mass of the object"... So yes this would imply on straight line motion also

7. May 14, 2015

### PeroK

If you forget about a rigid body for a moment. Consider a particle moving upwards in a straight line, starting a distance $a$ to the right of the origin. The angular momentum about the origin has magnitude $mav$ where $v$ is the speed.

If the particle is accelerating, then the angular momentum will be increasing and there will be angular acceleration. But, there is no circular motion involved.

The same is true for a rigid body. If one end starts at the origin and the rod accelerates up, the rod will have angular momentum and acceleration about the origin. Note that this is about the origin, not about the moving end of the rod. So, there is no rigid body rotation, but angular acceleration nevertheless.

Last edited: May 14, 2015
8. May 14, 2015

### Zalajbeg

I think this clarifies my thoughts. Just one more question:

If I know a body does only linear motion (let us say the rod I mentioned) but I don't know the magnitude of one of the forces. Then should I use the center of mass as the reference point for the torque balance? As I understand from your explanations the angular momentum regarding center of mass is zero if there is only linear motion and it is the only point that is valid. Therefore if I know a body is in linear motion I should check the torque balance regarding center of mass.

9. May 14, 2015

### PeroK

No point of reference is more "valid" than another. You choose your reference point to ease solving the problem. Moreover, all rigid body motion can be expressed as a combination of translation of centre of mass and rotation about centre of mass. For a free rigid body, this means it's usually best to consider torque about centre of mass.

If a rigid body is pivoted, it's usually best to consider torque about the pivot point.

With a gyroscope, for example, it's quite enlightening to consider torque about different points and check that the same motion is predicted.

10. May 14, 2015

### Zalajbeg

What I meant with valid was that: The only point which has zero angular momentum for a body which has only translational motion is the center of mass, this equation is not valid for any other point. I agree with your statement there is no more valid point, the validity I mentioned was the equation. Thanks for your helps. I will make my calculations regarding center of mass.