# Rotation of a Bar

1. Jul 11, 2012

### mattsap

Hello,

So, I have a bar with a sphere on each end. Mass is 1 for all objects.

I apply a force on one of the spheres. What force is the sphere on the other end feel?

What happens when I apply a force on both ends? Same direction? Opposite direction?

My thoughts:

If applied in the same direction, a translation would occur. However, suppose we are in a xyz plane. and a force of (1,0,0) was applied on one end--A, and (2,0,0) was applied on the other--B. I think A would experience a translation and rotation.

What are the equations to describe this?

-Matt =)

2. Jul 12, 2012

### Simon Bridge

You also need to specify length of the bar. Other dimensions are useful too.
En-ee-way...

If a single force is applied a distance from the center of mass, then the component of that force in the direction of the center of mass accelerates the center of mass by F=ma and the perpendicular component acts as an applied torque about the center of mass.

The "other end" will, thus, accelerate too ... the force is transmitted through the structure of the object.

3. Jul 12, 2012

### jbriggs444

The question is better posed as:

I apply a force to the sphere at one end, what acceleration would the other sphere undergo?

If you (for instance) apply a force of one kilogram at one of the spheres at right angles to the length of the bar then you have a knowable torque about the center of mass of the system. If you can compute the moment of inertia of the system then you can use:

angular-acceleration = d(rotation-rate)/dt = torque / moment-of-inertia

And if you kow the mass of the system then you can use

linear-acceleration = d(velocity)/dt = force / mass

Now your challenge is to compute the moment of inertia of the system.

The moment of inertia of a rod (or bar) of negligible thickness about its center is...
1/12 M L^2
where M is the mass of the rod and L is its length

The moment of inertia of a mass of negligible size is...
M r^2 = 1/4 M L^2
where r is the distance from the center and L is (in this case) the length of the bar

If your spheres are of non-negligible size then they also contribute to the moment of inertia
2/5 M r^2
where M is the mass of a sphere and r is the radius of a sphere

Let us assume that the spheres are of negligible radius. Then the moment of inertia of the system is:

1/12 M L^2 + 1/4 M L^2 + 1/4 M L^2 = 7/12 M L^2

M is given as 1 (kilogram, I suppose)
L is unknown and let's assume that it is expressed in meters).

So that's 7/12 L^2 in units of kg m^2

If you have a hypothetical force of 1 Newton applied at right angles to one of the spheres, that would result in a torque of L/2 Newton meters.

Dividing by the moment of inertia that gives an angular acceleration of...

L/2 / ( 7/12 L^2) = 24/14 L radians/sec/sec

Multiplying by the moment arm, that's 24/(14L) * L/2 = 24/28 meters/sec/sec in a direction opposite to that of the applied force.

Contrary to the Simon's assertion, you do NOT need to know the length of the bar. It cancels out [if the radius of the spheres and is small by comparison].

Now, that was just rotation. We still need to add in the ordinary linear acceleration. That's 1 Newton / 3 kg = 1/3 meters/sec/sec in the same direction as the applied force.

So the sphere on the opposite end experiences a net acceleration of 24/28 - 1/3 ~= 0.5 meters/sec in a direction opposite to that of the applied force.

I've probably muffed the math somewhere, but the general idea should be accurate.

[Editted to more carefully parenthesize under the division sign]

Last edited: Jul 12, 2012
4. Jul 12, 2012

### Simon Bridge

<checks> Oh I did say "need" didn't I? Oh well - depends what you want to do.
Important - I really just wanted to indicate that the geometry is important to the results and blew it.

In general it is useful to have the position vector from the center of mass to the applied force as well as the force vector. The question is a little on the general side in phrasing for the conclusions that were being drawn... but we do what we can.

Neither of us tackled what would happen for multiple forces at different places - which is, of course, that they all add up. We'd normally analyze this by adding the rotational and linear terms separately.

Two forces can cooperate to create just rotation or just translation if their linear or rotational components cancel respectively.

5. Jul 12, 2012

### mattsap

Well. This is a simulation. So the spheres are just point particles; while the rod is just an imaginary line connecting the two points. The unit can just be generalized as units. The distance between points is 1 unit length.

The only thing that matters in the simulation is the initial position and the final position (displacement) after all forces have been calculated.

Since we're given just a state and asked to find the next state. We're working in a 3D plane.

Does the direction of the force matter to determine translation or rotation?

6. Jul 12, 2012

### jbriggs444

This new problem specification seems to differ from the original.

In the original, "mass is 1 for all objects".
In the new, "the rod is just an imaginary line".

So which is it? Does the rod have mass 1? Or is the rod massless?

Now, you say that you are given a state and asked for the next state. That sounds like you have a discrete simulation for a continuous process. Does the term "Runge Kutta" mean anything to you?

Anyway, In answer to your question: "Does the direction of the force matter", the answer is "YES!!!"

7. Jul 12, 2012

### mattsap

In the new, "the rod is just an imaginary line".
This is more accurate. The reason for stating M is 1 was to try to say the mass is not being modeled in this simulation. But, I think Mass may or may not need to be considered.

"discrete simulation for a continuous process"
Exactly

"Does the term "Runge Kutta" mean anything to you?"
Yes.

8. Jul 12, 2012

### jbriggs444

Fair enough.

Note that I mentioned above that I might have muffed the math. I did.

Dividing a torque of L/2 by moment of inertia of 7/12 L^2 does not yield a rotational acceleration of 24/14 rad/sec/sec. It actually yields 12/14 / L rad/sec/sec.

With a massless connecting rod, that becomes L/2 divided by 1/2 L^2 = 1/L rad/sec/sec

The retrograde motion of the opposite sphere due to rotation then becomes -0.5 m/sec/sec
The forward motion of the opposite sphere due to linear acceleration is 0.5 m/sec/sec

The near-term net motion of the opposite sphere is zero (as one might expect by inspection).