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Concept of circular motion

  • Thread starter nil1996
  • Start date
301
7
1. The problem statement, all variables and given/known data
Hello PF:smile:,
i have a serious problem on circular motion.
My textbook says that
angular velocity =linear velocity /radius
While the my another book specialized in physics says that
angular velocity=vertical component of velocity/radius.

I am not understanding if both the book are saying the same.I am in agree with the solution given in the later book.
Is linear velocity same as vertical component of velocity?
 
Last edited:
1,540
133
1. The problem statement, all variables and given/known data
Hello PF:smile:,
i have a serious problem on circular motion.
My text says that
angular velocity =linear velocity /radius
While the my another book specialized in physics says that
angular velocity=vertical component of velocity/radius.

I am not understanding if both the book are saying the same.I am in agree with the solution given in the later book.
Is linear velocity same as vertical component of velocity?
It should be tangential component of velocity. "vertical component of velocity" doesn't make any sense
 
301
7
It should be tangential component of velocity. "vertical component of velocity" doesn't make any sense
Here i go confused:confused:. isn't velocity already tangent to the curve?
 

haruspex

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It should be tangential component of velocity. "vertical component of velocity" doesn't make any sense
One small quibble - it does say velocity, not speed. Velocity implies vectors, and you cannot divide by a vector.
In vectorial terms, ω = r x v/r.r. Since that's the cross product, it only depends on the component of the velocity that's orthogonal to the radius.
 
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133
Here i go confused:confused:. isn't velocity already tangent to the curve?
No need to get confused :smile:.Look at the picture I have attached.

Let the particle be at P with velocity vector in red .The point about which the angular velocity is calculated is O .OP is the position vector .The component of velocity along OP is in green (not useful for calculating angular velocity) .The component of velocity perpendicular to OP is in blue (Let us represent its magnitude by vT)

Now,angular velocity of P with respect to O = vT/OP
 

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301
7
So as concerned to the circular motion both the books are right.
Thanks for that explanation.
 
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Here i go confused:confused:. isn't velocity already tangent to the curve?
Of course it is. The linear velocity vector is tangent to the trajectory.
The confusion comes from the fact that in some textbooks they use "tangential" velocity to differentiate it from "angular" velocity.
I think that is better to use "linear" velocity and "angular" velocity. Like in your first example.
The linear velocity is tangent to the trajectory so if you call it tangential velocity is not untrue but can be confusing.

On the other hand, what Tanya shows in green as the component that matters for calculating the angular acceleration is perpendicular to the position vector from O but it is not the tangential component, as tangential refers usually to "tangent to the trajectory".

So I think that in general none of the definitions is "right". Even overlooking the problem with vector division (assuming that they mean magnitudes of the vectors).
 
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One small quibble - it does say velocity, not speed. Velocity implies vectors, and you cannot divide by a vector.
In vectorial terms, ω = r x v/r.r. Since that's the cross product, it only depends on the component of the velocity that's orthogonal to the radius.
haruspex...Thanks for the quibble :)
 
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nil1996...

I apologize for using an incorrect term "tangential component of velocity" in post#2 .Instead, please read it as "component of velocity orthogonal(perpendicular) to the position vector(radius) of the particle with respect to the origin".
 
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thanks got it right :)
 

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