Circular motion doubt: Angular velocity vector for general planar motion about a point in the plane

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The discussion revolves around the relationship between angular velocity, linear velocity, and the radius vector in general planar motion. It clarifies that the linear velocity vector cannot be expressed as the angular velocity vector multiplied by the radius vector, which is deemed false. The correct interpretation involves recognizing that the tangential velocity is derived from the radius and angular displacement, leading to the equation vθ = r(dθ/dt). Participants express confusion about how changes in radius affect tangential velocity, but ultimately reach an understanding that tangential velocity is indeed r times angular velocity. The conversation concludes with a resolution of the initial doubts regarding the physical interpretation of these relationships.
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Homework Statement
Angular velocity vector = f( linear velocity vector , radius vector) for a general planar motion about a point in the plane. Suggest true or FALSE
a) linear velocity vector = angular velocity vector × radius vector.
b) Perpendicular velocity vectors= angular velocity vector × radius vector
Relevant Equations
The answer for the first one is given false and the answer to the second one is given as true.
I thought the opposite should be true since its a general planar motion its not necessary that the magnitude of radius vector is constant so the change in direction and magnitude of radius vector should be generated by the perpendicular velocity vector and parallel velocity vector respectively ? Therefore the whole velocity vector should be taken instead of only the perpendicular component since that Is only responsible for changing direction.... It does make sense mathematically since cross product creates another vector perpendicular to the two vectors taken but I cannot make sense of it physically...
 
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tellmesomething said:
Homework Statement: Angular velocity vector = f( linear velocity vector , radius vector) for a general planar motion about a point in the plane. Suggest true or FALSE
a) linear velocity vector = angular velocity vector × radius vector.
b) Perpendicular velocity vectors= angular velocity vector × radius vector
Relevant Equations: The answer for the first one is given false and the answer to the second one is given as true.

I thought the opposite should be true since its a general planar motion its not necessary that the magnitude of radius vector is constant so the change in direction and magnitude of radius vector should be generated by the perpendicular velocity vector and parallel velocity vector respectively ? Therefore the whole velocity vector should be taken instead of only the perpendicular component since that Is only responsible for changing direction.... It does make sense mathematically since cross product creates another vector perpendicular to the two vectors taken but I cannot make sense of it physically...
Consider a radial displacement ##dr## and an angular displacement ##d\theta##, leading to a tangential displacement ##rd\theta##, in time ##dt##. Taking the limit, ##v_r=\frac{dr}{dt}, v_{\theta}=r\frac{d\theta}{dt}##.
E.g., consider the case of a purely radial velocity.
 
haruspex said:
Consider a radial displacement ##dr## and an angular displacement ##d\theta##, leading to a tangential displacement ##rd\theta##, in time ##dt##. Taking the limit, ##v_r=\frac{dr}{dt}, v_{\theta}=r\frac{d\theta}{dt}##.
E.g., consider the case of a purely radial velocity.
I am not sure I understand....you say that radius times angular velocity = velocity theta...is velocity theta the perpendicular component of velocity? If so why?
 
tellmesomething said:
I am not sure I understand....you say that radius times angular velocity = velocity theta...is velocity theta the perpendicular component of velocity? If so why?
By ##v_{\theta}## I meant the velocity in the tangential direction. Do you understand that the displacement in the tangential direction is ##rd\theta##?
 
tellmesomething said:
Yes, such a small displacement becomes equal to the arc length i.e distance so ds=rd(theta). I get that, but I dont get how you get tangential velocity=rd (theta)/dt.
Just divide both sides by dt:
##ds=rd\theta##
##\frac{ds}{dt}=r\frac{d\theta}{dt}##
where ##ds## is the tangential displacement.
 
haruspex said:
Just divide both sides by dt:
##ds=rd\theta##
##\frac{ds}{dt}=r\frac{d\theta}{dt}##.
Yes sorry got it. After this we get tangential velocity = r d(theta)/dt but how is this accounting for the change in the radius?
 
haruspex said:
Just divide both sides by dt:
##ds=rd\theta##
##\frac{ds}{dt}=r\frac{d\theta}{dt}##
where ##ds## is the tangential displacement.
That's silly of me, it all makes sense sorry for the trivial overlook, thankyou I got it. r * d(theta)/dt is nothing but r times angular velocity. Thankyou again
 
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