How to show that a particle with position vector moves on a circular path, has constant speed, and its...
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How to show that a particle with position vector moves on a circular path, has constant speed, and its...

[From: ] [author: ] [Date: 11-10-04] [Hit: ]
Thanks for any insight!-Consider the distance of r from the origin at time t.= sqrt ([A]^2),= A, so the particle is always a constant distance A away from the origin, it moves in a circular path with radius A.......
...velocity is tangent to the circle? The equation for the position vector is r = Asin(wt)i + Acos(wt)j, and a and w are constants. Thanks for any insight!

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Consider the distance of r from the origin at time t.
This can be reprensented as:
sqrt( [Asin(wt)]^2 + [Acos(wt)]^2 )
= sqrt( [A]^2 * ([sin(wt)]^2 + [cos(wt)]^2))
= sqrt ([A]^2), {because sin^2 + cos^2 = 1}
= A, so the particle is always a constant distance A away from the origin, it moves in a circular path with radius A.
Note this position vector r points in the direction of the radius from the origin to some point on the circle at some time t.

Its velocity vector will be v = dr/dt = wAcos(wt)i - wAsin(wt)j
The magnitude of this gives you the speed:
magnitude (v) = sqrt( [wAcos(wt)]^2 + [-wAsin(wt)]^2 )
= sqrt( [wA]^2 * ([cos(wt)]^2 + [sin(wt)]^2))
= sqrt ([wA]^2)
= wA, which is a constant, so speed is constant.

To prove velocity vector v is tangent to the circle, we use the dot product of v and r.
(velocity vector and position vector). If the dot product of two vectors is zero, the two are perpendicular.
v dot r = ( wAcos(wt)i - wAsin(wt)j ) dot ( Asin(wt)i + Acos(wt)j )
= [wAcos(wt)]*[Asin(wt)] + [- wAsin(wt)]*[Acos(wt)]
= w(A^2)*(sin(wt))*(cos(wt)) - w(A^2)*(sin(wt))*(cos(wt))
= 0
Therefore v and r are perpendicular at any time t, since r points in the direction of the radius of the circle, v is perpendicular and is tangent to the circle.
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