# Concept of circular motion

1. Nov 15, 2013

### nil1996

1. The problem statement, all variables and given/known data
Hello PF,
i have a serious problem on circular motion.
My textbook says that
While the my another book specialized in physics says that

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: Nov 15, 2013
2. Nov 15, 2013

### Tanya Sharma

It should be tangential component of velocity. "vertical component of velocity" doesn't make any sense

3. Nov 15, 2013

### nil1996

Here i go confused. isn't velocity already tangent to the curve?

4. Nov 15, 2013

### haruspex

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.

5. Nov 15, 2013

### Tanya Sharma

No need to get confused .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

#### Attached Files:

• ###### velocity.PNG
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6. Nov 15, 2013

### nil1996

So as concerned to the circular motion both the books are right.
Thanks for that explanation.

7. Nov 15, 2013

### nasu

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).

Last edited: Nov 15, 2013
8. Nov 15, 2013

### Tanya Sharma

haruspex...Thanks for the quibble :)

9. Nov 15, 2013

### Tanya Sharma

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".

Last edited: Nov 15, 2013
10. Nov 21, 2013

### nil1996

thanks got it right :)