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#1 2009-04-22 01:16:05

Onyx
Member
Registered: 2009-02-24
Posts: 48

moment of vector about point in 2D

Hi, given a vector and point in R^2, how do we find the moment of the vector about the point? e.g say we have the vector <3,1> at position (-2,5), and we want to find its moment about the origin.

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#2 2009-04-22 03:55:46

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: moment of vector about point in 2D

a force F, applied at a point P on a body which forms the radial vector R from the centre of rotation, to the point P, will have a moment of:

M = R×F

this is in 3D, in 2D it degeneralises to:

m = RyFx - RxFy

since you are asking the moment about the origin, the radial vector R is simply P

in your example this is:

m = 5*3 + 2*1 = 15 + 2 = 17

Last edited by luca-deltodesco (2009-04-22 03:58:34)


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#3 2009-04-23 02:43:56

Onyx
Member
Registered: 2009-02-24
Posts: 48

Re: moment of vector about point in 2D

Thanks, I had seen the cross product used in 3D examples, and know this isn't defined in 2D, but I see the formula you've given is that of a determinant. Basically we're finding the component of the force vector acting perpendicuar to the position vector of the point right? (assuming the moment is about the origin)

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#4 2009-04-23 03:58:46

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: moment of vector about point in 2D

pretty much yeh, only that is it also multipled by the distance from the point being rotated by

evaluating that 'determinant' as you call it (whilst true, shouldn't really be calling it that) the 'perp-dot product' follows the same rules as the cross product in 3d: namely that:

thus evaluating the perp-dot gives you the distance from the axis of rotation, multiplied by the magnitude of the force, and the sine of the angle between the two vectors, which gives you the magnitude of the force perpendicular to the radial vector, multiplied by the distance from the axis

Last edited by luca-deltodesco (2009-04-23 04:01:10)


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