Prove v^ has all of a vector's quantities

[TheColector]

Homework Statement

Hi
Given the linear velocity formula: v* v^ = r*ω(-sinθi^ + cosθj^)
i^, j^, v^ - unit vectors
I'm to prove that v^ has direction, turn and magnitude

Magnitude:
|v^| = sqrt((-sinθ)^2 + (cosθ)^2) = 1 (as is also stated in unit vector's definition)

Direction and turn:
-sinθ i^ +cosθj^ - describes a circular motion
And this is the part I can't prove. How am I to prove that formula above describes circular motion ? And how to determine direction and turn ?( I mean it changes every time the movement is made)
I can explain it on a paper with a drawing of a circle and the object moving there.

 

[kuruman]

First off you got the magnitude wrong. What you proved to be 1 is the magnitude of the unit vector not the magnitude of the velocity, a.k.a. the speed.
Secondly, to show circular motion, compare the tip of the velocity vector of an object going around a circle with the tip of the velocity vector that is given to you as θ varies from zero to 2π.

 

[haruspex]

What you proved to be 1 is the magnitude of the unit vector

According to the question statement, that is what was of interest, though I assume the task is to find direction, turn and magnitude, not to prove that $\hat v$ has them.

 

[haruspex]

How am I to prove that formula above describes circular motion ?

You could consider radial velocity.

 

[kuruman]

According to the question statement, that is what was of interest, ...

I guess I was confused by the unorthodox notation. If quantity r is starred, then ω should also be starred as part of the magnitude v^* = r^*ω^*.

 

[TheColector]

It was indeed about calculating magnitude of the unit vector.
In order to use radial velocity I assume that given motion is a circular one
I might have actually mislead you in the description. I'm to prove that v^ has magnitude(which I did) , turn and direction)
The thing is I have no idea why. The approach I have just made is as follows:
- let's take x to be [r*ω*sin(θ)] and y to be [r*ω*cos(θ)]
- then when we take x^2 + y^2 = r^2 *ω^2 - and this is a formula for a circle equasion
Neither it is wrong nor right according to my logic. What do you think ?
That would just explain direction and turn as it is a circular motion.

 

[kuruman]

x^2 + y^2 = r^2 *ω^2

More correctly v_x² + v_y² = r²ω². That is indeed the equation of a circle if r ω is constant. Assume that it is and consider the velocity vector when θ has different values, say 0, π/2, π, 3π/2 and 2π. Draw it and compare with the velocity vector of an object that is going around a circle counterclockwise when it is at the 3 , 12, 9 and 6 o'clock positions.

 

[TheColector]

@kuruman Yeah that seems to be correct. Thanks for your help.
Take care

 

[haruspex]

In order to use radial velocity I assume that given motion is a circular one

No, I thought you wanted to prove it was circular.
If you take the dot product of the velocity vector with the position vector then you will find if the radial velocity is zero. If it is, it must be circular motion.