Combining linear and rotational equations of motion

[JS1103]

I have a moving body with constant linear and rotational acceleration. Given a starting condition (position, orientation, linear and angular velocities), how can I combine the equations of motion to give a position and orientation a given time on?

 

[mfb]

You can calculate the displacements from both effects separately and add them.
If that does not help, it would be interesting to know more about the setup you consider.

 

[JS1103]

Thank you for your reply mfb, however, I am struggling to get the results I'm looking for.

As a simplified example, if we have an initial position of (0, 0) m, orientation of 0 rad, velocity of (1, 0) m/s, angular velocity of π/60 rad/s and zero linear and angular acceleration, then after 60 s:
r = r₀ + v₀t + 0.5at² = (60, 0) m
θ = θ₀ + ω₀t + 0.5αt² = π rad

So calculating separately and combining, we have traveled 60 m in a straight line and are now facing back to our start point.
What I was intending is that we have traced an arc with the position being (0, 120/π) m.

 

[A.T.]

zero linear and angular acceleration ... What I was intending is that we have traced an arc

You can't trace an arc with zero linear acceleration. You want linear velocity coupled to orientation, so it changes direction, which constitutes a linear acceleration.

 

[JS1103]

Could you please explain how I can achieve this? In my example above, what formula would I use to calculate the position at any given time if the body is rotating and the velocities are in the body's axes?

 

[CWatters]

"initial position of (0, 0) m"
"velocity of (1, 0) m/s"
"zero linear ... acceleration"

If it has zero velocity and zero acceleration in the y-axis the displacement in the y-axis will remain unchanged no matter what happens in x or orientation. So it must end up at (something, 0) .

Circular motion in Cartesian coordinates involves terms like...

y = RCos(wt)
x = RSin(wt)

where R is the radius (60/π)
w = angular velocity (in this case you want to travel 180 degrees around the arc in 60 seconds = π/60)
t = time (0-60)

Something like that anyway.

 

[JS1103]

If it has zero velocity and zero acceleration in the y-axis the displacement in the y-axis will remain unchanged no matter what happens in x or orientation. So it must end up at (something, 0)

The velocity is in the body's axes, so given it is rotating, why must it end up at (something, 0)?

 

[nasu]

Thank you for your reply mfb, however, I am struggling to get the results I'm looking for.

As a simplified example, if we have an initial position of (0, 0) m, orientation of 0 rad, velocity of (1, 0) m/s, angular velocity of π/60 rad/s and zero linear and angular acceleration, then after 60 s:
r = r₀ + v₀t + 0.5at² = (60, 0) m
θ = θ₀ + ω₀t + 0.5αt² = π rad

So calculating separately and combining, we have traveled 60 m in a straight line and are now facing back to our start point.
What I was intending is that we have traced an arc with the position being (0, 120/π) m.

The position of what is like that?

You should be aware that the nice decomposition into translation and rotation works if you take the translation of the center of mass and the rotation around the center of mass.
Are your initial conditions (linear velocity and position) referring to the center of mass?

 

[JS1103]

The position is the body's centre of mass in a fixed axis system. The linear velocity and acceleration is in the body's axes. I would like to be able to calculate the body's location in the fixed axis system and the velocity in the body's axes at any given time, assuming the linear and rotational acceleration are constant.

 

[mfb]

Rotation around what? If it is around the center of mass, and if we consider the center of mass only here, then rotation does not influence the position at all.
If you consider a point somewhere else on the object, then your position calculation is wrong, and the point will describe an arc.

 

[JS1103]

I'll try an illustration to help explain what I'm trying to do.

In all three scenarios above I have plotted the position over 60 s starting from (0, 0) with an initial orientation along the x-axis and no linear acceleration (in the body's axes).
In Scenario 1, I have an initial velocity of (1, 1) m/s, and no rotational velocity or acceleration.
In Scenario 2, I have also introduced an initial angular velocity of π/60 rad/s. The velocity remains as (1, 1) m/s relative to the body, but as the body's orientation is changing, it traces an arc.
In Scenario 3, I have also introduced an angular acceleration of π/600 rad/s², which then causes it to spiral.

I would like to know, given a constant acceleration in the body's axes, how do I calculate its position and velocity in the fixed axes at any given time?
I appreciate my original question was not precise enough. Thank you for your help so far in understanding what I was actually trying to ask!

 

[CWatters]

The velocity is in the body's axes, so given it is rotating, why must it end up at (something, 0)?

Sorry I didn't see your post #5 when I posted #6.

 

[nasu]

The position is the body's centre of mass in a fixed axis system. The linear velocity and acceleration is in the body's axes. I would like to be able to calculate the body's location in the fixed axis system and the velocity in the body's axes at any given time, assuming the linear and rotational acceleration are constant.

The linear velocity relative to the body's axes? Linear velocity of what? You mean a specific point of the body?
Using the body's axes is tricky. You consider these axes as moving with the body? Or fixed? Or moving but only by translation?

 

[CWatters]

In Scenario 2, I have also introduced an initial angular velocity of π/60 rad/s. The velocity remains as (1, 1) m/s relative to the body, but as the body's orientation is changing, it traces an arc.

Ok but that's not how physical objects behave in the real world. Bodies have momentum so simply rotating them does not alter the direction component of their velocity. For that you need an acceleration.

Edit: If you throw a ball straight up it will go straight up regardless of any spin you give it.

 

[nasu]

Something is unclear here.
You assume that the points of the body move relative to the body? Isn't this a rigid body?

 

[CWatters]

or put it this way... You say...

The velocity remains as (1, 1) m/s relative to the body

How can a body can have a velocity relative to itself?

 

[CWatters]

Perhaps look up centripetal acceleration. If the object moves in a circle that implies it's being accelerated towards the centre of the circle eg at right angles to it's instantaneous direction of motion. It doesn't have velocity in that direction.

 

[JS1103]

Thank you for all the replies.

Using the body's axes is tricky. You consider these axes as moving with the body? Or fixed? Or moving but only by translation?

Yes, the axes are moving with the body, following its orientation.

Edit: If you throw a ball straight up it will go straight up regardless of any spin you give it.

It produces its own thrust, more like a rocket than a ball, so it should move in a different direction as it spins.

Perhaps look up centripetal acceleration

Thank you. I will look into this.

 

[CWatters]

It produces its own thrust, more like a rocket than a ball, so it should move in a different direction as it spins.

That makes more sense. To move in a circle start with some initial velocity (v) and generate a sideways thrust (force) perpendicular to v (eg pointed at the centre of the circle). This will generate centripetal acceleration. The required force = mv²/r where m=mass of object, v = velocity, r = radius.

However this assumes you are modelling mass and inertia correctly. I suspect you might not be.