link to image When you apply a force, F (red) to the disk, it will cause a torque, T (blue), and a linear force, ? (green). How do you calculate the leftover linear force (green)? It can''t be the same as the initial force, because some force is exerted in the form of torque, right? Thanks, I hope I''m not too confusing.

Neglecting any frictional stuff I would hazard splitting the force up into a component tangent to the circle at the point of impact, and another compenent perpendicular to it. The former goes into torque, the latter into linear acceleration.

Actually, the direction of the resulting rotation would be counterclockwise, the opposite of the blue arrow you''ve drawn. Force F is tilted ever so slightly downward, and if you imagine an exaggeration of the tilt (like a force pointed down the left side of the wheel), the wheel would move in a counterclockwise fashion.

Torque is defined as the magnitude of the force multiplied by the perpendicular distance from the point of application to the axis of rotation, in this case the center of the wheel. You''d multiply only the component of force F perpendicular to the radius by the radius. If the perpendicular component points down the left side of the wheel, the resulting torque would lead to a counterclockwise rotation. The then "leftover" force would directly translate into the linear motion of the wheel.

RapscallionGL - arriving soon.

Ahh, so just use the leftover force? That is easy enough.

Thanks.

[EDIT]

The resultant linear velocity would keep the same direction as the initial Force vector, right?

[/EDIT]

[edited by - rpgman on October 28, 2002 12:14:45 AM]

The resultant linear acceleration would be in the direction of the force vector, yes.

RapscallionGL - arriving soon.

quote:

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Actually, the direction of the resulting rotation would be counterclockwise, the opposite of the blue arrow you've drawn. Force F is tilted ever so slightly downward, and if you imagine an exaggeration of the tilt (like a force pointed down the left side of the wheel), the wheel would move in a counterclockwise fashion.

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I've never been too good at these, but I believe you are wrong . To determine which way the wheel will rotate, you would draw a normal to the wheel at the point of contact, and compare that with the force being applied.

You say that the force is tilted "downward", but if you took the entire diagram and rotated clockwise it about the centre of the circle so that the force was horizontal, then the information is still exactly the same, but the force is no longer "tilted" at all. The tilt you mention is purely relative to the image, and nothing else.

I think for the actual calculations, you need to firstly work out how much of the force is tangential to the wheel. Eg:

If the force is a (5, 1) and the (normalised) tangent vector is (1, 0), then the entire horizontal component of the force will be torque. What is left over is the vertical component, thus the linear force will actually be (0, 1) which is obviously not in the same direction as the force vector.

Trying is the first step towards failure.

[edited by - ragonastick on October 28, 2002 12:35:59 AM]

quote:

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Original post by RPGman
The resultant linear velocity would keep the same direction as the initial Force vector, right?

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If I understand the question correctly, you ask if the red line in your original diagram would have the same direction as the green line (parallel).

I''m not sure it would. I''m just logically thinking of pool (billiards) balls. When the cue ball strikes the object ball straight on, the object ball goes straight. But when you dont hit the object ball straight on, the object ball will go off at an angle.

That angle, I believe, is the inverse of the vector from the center of the disk to the point of impact on the rim of the disk.

Actually, now I think some more, I think that perhaps that the torque force is not necessarily the maximum bit of the tangential force. Just playing with some more examples in my mind, consider:

Force is horizontal: (5, 0)
Normalised Tangential Vector (a, a) (a = (Sqr(2) / 2))

This is *fairly* similar to the diagram.

Now, because the tangent has both a vertical and horizontal component, we can''t just ignore one component of the force like I did in earlier examples.

New thinking:

If you took the dot product of the two vectors, (with the tangent normalised and the force not) it will give you the force which goes into radial stuff, so then you multiply it by the radius to get the torque. Now, scale the tangential vector by the newly found dot product and subtract this vector from the initial force to get the linear force.

I am writing that without anything other than intuition - write a demo to see if it looks correct becasue I''m just going with my gut on this one. When F is in the same direction as the Tangent, then the dot product is |F| thus the force will just make the wheel spin with no linear movement. If they are perpendicular and the force is going directly into the wheel, then the dot product will be 0, thus there is no torque and all movement is linear. Hopefully all the in between cases are correct too

I think this is a pretty difficult problem Obviously we can split the force up into two parts. One is tangent to the surface of the sphere - call it F_T and the other F_P is perpendicular to it. All of F_P goes into linear acceleration since cannot apply any torque to the object.

Part of F_T will provide an angular acceleration CLOCKWISE, and part of it will provide a linear acceleration of the body as a whole. What proportion of the force does what requires further information.

Torque = |F| * |R| * sin( theta ), where R is the vector from the center-of-mass to the point where F is being applied, and theta is the angle between F and R (about 135 degrees in your picture).

As for how to calculate the force on the center of mass (your green arrow): Intuitively, you might think that force and kinetic energy should be "conserved". So a force that produces a lot of torque should "compensate" by producing less linear acceleration. This intuition is wrong. The force that you should apply to your center of mass (your green arrow) is actually just the full, original force (your red arrow). If this seems illogical to you, maybe this rant will help:

What you're actually conserving in these situations is *work* and kinetic energy. For example, pushing on a door knob for 1 second accelerates the door a lot, but pushing on the door near the hinges for 1 second will barely accelerate the door at all. Here we see that equal forces have produced unequal changes in the kinetic energy of the door. The key is that, in the former case, your hand is pushing over a distance of a few feet, whereas in the latter case, your hand is pushing over a distance of a few inches. Work = force * distance, so you're doing a lot more work in the former case.

For a much better explanation of this stuff, try Chris Hecker's physics articles:
http://www.d6.com/users/checker/dynamics.htm

[edited by - Eric on October 29, 2002 4:35:19 AM]