# Pushing on the side of a sphere, for torque do I use the tangential force?

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I'm adding a thruster to the side of a sphere so when fired it will cause some angular and linear movement. I'm guessing I should split force of this pushing to the normal and tangential forces, so the normal one will add linear movement to the sphere, and the tangential(originalForce - normalForce) will create some torque, is this correct? Also I'm trying to work out how to calculate the position, orientation, linear and angular velocities at a certain time, when a thruster is being fired the whole time. My major problem is when the accleration imparts rotation, because this would cause the linear and angular accleration to be applied in various directions over time. Any idea on how I would go about that? thanks.

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The amount of torque on an object is the cross product of the force's offset from the object's center of mass and the force.

t = r X F

If that helps..

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Torque would be the radius of the sphere times the tangential force, though you may find it easier to just use one force vector and calculate torque as the cross product of the force and the radius vector (vector from the center of the sphere to the point where the force is being applied).

The kinematic equations for position, velocity and acceleration are all the same. Acceleration equals force over mass. Equations for rotation are analogous to the linear equations, ie: Angular acceleration equals torque over the moment of inertia. Angular velocity equals angular acceleration times time, etc, you get the picture.

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I'm confused about the linear movement part.

Would a sphere get any linear accleration if I attached a thruster to the sphere with its direction perpendicular to the centre of the sphere? (from a top down point of view).

Because it feels like it should get some linear movement from the thrust, but when I think about the maths, it seems like there would be zero linear accleration (eg dot product at 90 degrees would equal 0 i believe).

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I had the same misconception as you at one point, so I know how you feel. All forces acting on an object, regardless of orientation, can be seen as acting on the objects center of mass from a translational point of view. There isn't a dot product involved in applying the force. Applying a force tangental to a sphere right on its surface would cause it to move forward in addition to spinning. If the source of the force is attached to the sphere, though, then the result will just be a wild loop-de-loop as the sphere rotates. Point is that translational and rotational frames of reference can be seen as entirely seperate and independant.

A case where your real world intuition trumps your mathematical intuition.

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When applying F to the sphere...

M = mass of sphere
I = moment of inertia of sphere
R = location where force is applied at
N = force direction * unit vector that points to center of mass
T = force direction cross unit vector that points to center of mass

amount of linear motion is N + T * R*T.unit()/(M + R*T.unit())
amount of rotational motion is T * M/(M + R*T.unit())

Well, my math isn't very precise or necessarily correct to the finest details, but the basic idea is that as the part of the force that pushes directly towards the center of the sphere increases, the linear motion of the sphere increases.

The tangential part of the force increasingly affects linear motion more than angular motion the closer that R is to the center of mass. So if you applied the force tangential to the center of mass of the sphere against a weightless rod that was infinitely long, then the sphere would gain no linear motion from that force. If the mass of the sphere was infinite, and the moment of inertia was finite, then any tangential force would also only cause a rotation. Otherwise, its a ratio of force applied to linear/angular momentum.

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Quote:
 Original post by pTymNWell, my math isn't very precise or necessarily correct to the finest details, but the basic idea is that as the part of the force that pushes directly towards the center of the sphere increases, the linear motion of the sphere increases.

I don't think this is correct, if I'm understand your point properly. Check out Chris Hecker's physics introduction, page 5 of 8, last paragraph. To quote:

Quote:
 In short, Eq. 11 tells us we can treat all the forces acting on our rigid body as if their vector sum is acting on a point at the center of mass with the mass of the entire body. We divide a force (say, gravity) by M to find the acceleration of the center of mass, and then we integrate that acceleration over time to get the velocity and position of our body... Note that Eq. 11 doesn’t contain any information about where the forces were applied to the body. That information drops out when dealing with linear momentum and the center of mass, and we just apply all forces to the CM to find its acceleration...

Bolds added for emphasis. It is, really, as simple as you were lead to believe from physics 101. F = m * a. No dot products, cross products, normals, etc. used.

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Yeah Numsgil is correct.

Edit:

I tried working out the motion for the 2D case (circle + thruster). The angular part is trivial but the center of mass motion is a little tricky. I'll try to run it numerically and post a picture.

[Edited by - My_Mind_Is_Going on June 18, 2007 1:26:29 AM]

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