Hi, how to find the magnitude of a rotating vector?

[rashida564]

i get stuck in how to find the magnitude of rotating vector . why say that |dA/dt|=A(dθ/dt) but who we can derive it or interpret this fact

 

[Simon Bridge]

Start from the definitions.

 

[rashida564]

the rotating vector have a constant magnitude but it change it's angle by the time

 

[Simon Bridge]

OK. So write that out mathematically.

 

[rashida564]

write the definition mathematically

 

[Simon Bridge]

Do you not know how to write down the equation for a rotating vector?
A vector in rectangular coordinates is written: ##\vec v = v_x\hat\imath + v_y\hat\jmath##
The magnitude of that vector is ##|\vec v| = v = \sqrt{v_x^2+v_y^2}##

If the vector makes angle ##\theta## to the x-axis, then:
##v_x=v\cos\theta,\; v_y=v\sin\theta##
If the vector rotates, then ##\theta## is a function of time.

From there, and the definition of the derivative, you can derive and interpret the relation yourself.
So here is your task: prove ##\big|\frac{d}{dt} \vec v(t) \big| = v\frac{d}{dt}\theta(t)##

 

[rashida564]

let v=r
dr/dt = lim Δt→0 Δr/Δt
Δr=2rsin(θ/2)
Δr≈rθ
dr/dt=lim Δt→0 Δrθ/Δt
|dt/dt| = r(dθ/dt)
is it correct

 

[Simon Bridge]

No. dt/dt = 1, your delta-r is a non sequitur... basically, what you wrote is nonsense.
It looks like you need to revise your notes about writing vectors.

 

[rashida564]

i saw the explanation from Kleppner and Kolenkow book

 

[Simon Bridge]

What explanation?
Reread post #6. If you will not take advise I cannot help you.
Do you know how to write a vector?

 

[rashida564]

a vector in rectangular coordinates system is written by it's x and y component
then we should write the vector as vcos(θ)i+vsin(θ) j

 

[Simon Bridge]

Well done... so write that out, A= or whatever letter you wsnt to use for a vector.
Then differentiate both sides with respect to time ... you will need the chain rule because ##\theta## is a function of time.

 

[rashida564]

dA/dt=-sin(θ)dθ/dt i +vcos(θ)dθ/dt j

 

[Svein]

dA/dt=-sin(θ)dθ/dt i +vcos(θ)dθ/dt j

You forgot the v in front of sin(θ). Now figure out what angle there is between A and dA/dt .

 

[Simon Bridge]

Don't worry about the angle... it's a shortcut but you don't need to figure that out.
Your next step is to find the magnatude. Don't forget that |A| = v
Find the expression for |dA/dt| and simplify.

 

[rashida564]

now i get it
|V|= √((-vsin(θ)dθ/dt)^2+(vcos(θ)dθ/dt)^2)
|V|=vdθ/dt*√(sin^2(θ)+cos^2(θ))
|V|=vdθ/dt

 

[Simon Bridge]

Well done.
Sometimes you have to just go through the steps without knowing for sure where you will end up.
Of course I could just have told you ... but then you'd have missed out on that "ahah" moment.

 

[rashida564]

thank you