What are the ##\vec{u}_r## and ##\vec{u}_θ## vectors?

[RyanH42]

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color]

In textbook I am troubling understanding the question.Actually I don't know the terms $\vec{u}_r$ and $\vec{u}_θ$
Here the question.
A particle moves according to the polar equation $r=1+cosθ$, $θ=e^t$ where t is in seconds are r in feet.What are the $\vec{u}_r$ and $\vec{u}_θ$ components of acceleration of this particle at the instant $t=ln(π/2)$.
I don't want you to answer the question.I write the question to clearify my question.
I know write accelaration vector in terms of $\vec{T}$ and $\vec{N}$ not $\vec{u}_θ$ and $\vec{u}_r$.If you can show me how to write it,I can do the question(Thats why I didnt open this thread in Homework section)
Thanks

 

[Vatsal Sanjay]

Actually I don't know the terms u⃗ r\vec{u}_r and u⃗ θ\vec{u}_θ

u_r is the vector along the radial velocity direction of the particle and u_$\theta$ is the vector along the tangential velocity direction.

I know write acceleration vector in terms of T⃗ \vec{T} and N⃗ \vec{N} not u⃗ θ\vec{u}_θ and u⃗ r\vec{u}_r.If you can show me how to write it,I can do the question(

when you will differentiate the radius vector (as given in polar coordinates); you will get two components of velocity; one along the radius vector itself, responsible for changing the magnitude of the vector and one perpendicular to the radius vector, responsible for changing the direction of radius vector. On further differentiation of the velocity vector, you will again get an expression for acceleration; a vector which will have components in the above-mentioned two directions.

P.S. For the cylindrical coordinate system; the two vectors you are looking for in this case are the radius vector along the radial direction and the tangential vector along the azimuthal direction.

 

[Chestermiller]

u_r is the vector along the radial velocity direction of the particle and u_$\theta$ is the vector along the tangential velocity direction.

This is not quite right. The tangential velocity is the component of the velocity vector tangent to the trajectory of the particle. This is not generally in the direction of u_$\theta$. u_$\theta$ is the unit vector in the θ direction.

Chet

 

[RyanH42]

So If I differantiate it respect to $t$ I will find speed vector.$d\vec{R}/dt=(d\vec{R}/dθ).(dθ/dt)=-sinθe^t$ and I will differantiate respet to $t$ again to find acceleration vector that's is .$d\vec{V}/dt=(d\vec{V}/dθ).(dθ/dt)=(-sinθ+-cosθθ).e^t$
Is it true so far ?

 

[Nathanael]

So If I differantiate it respect to $t$ I will find speed vector.$d\vec{R}/dt=(d\vec{R}/dθ).(dθ/dt)=-sinθe^t$

The time derivative of a vector should be another vector, right? I'm guessing you meant to say d\vec{R}/dt=(-sinθe^t)\vec u_r right?

Anyway this is not correct because you're assuming \vec u_r is a constant with respect to time. This is an important difference between polar and cartesian coordinates: cartesian unit vectors point in a constant direction whereas polar unit vectors point in a direction that depends on θ. (And so if θ changes with time, the unit vectors will change with time.)

You have to use the product rule: \frac{d}{dt}(r\vec u_r)=\frac{dr}{dt}\vec u_r +r\frac{d\vec u_r}{dt}

I will leave it to you to try and figure out what \frac{d\vec u_r}{dt} is equal to.

 

[RyanH42]

$\vec{R}=r\vec{u}_r$ then as you said $v=(dr/dt)\vec{u}_r+r(d\vec{u}_r/dt)$
so its $\vec{v}= (-sinθe^t)\vec{u}_r+(1+cosθ)e^t\vec{u}_θ.$
so acceleration vector.$\vec{a}=(d^2r/dt^2)\vec{u}_r+dr/dt(d\vec{u}_r/dt)+dr/dt(d\vec{u}_r/dt)+r(d^2\vec{u}_r/dt^2)$
.$\vec{a}=((-sinθ+-cosθθ).e^t)\vec{u}_r+2(-sinθe^t).(\vec{u}_θ.e^t)+(1+cosθ).(-\vec{u}_re^t+\vec{u}_θ)e^t$

Is this true ?

 

[Nathanael]

$\vec{R}=r\vec{u}_r$ then as you said $v=(dr/dt)\vec{u}_r+r(d\vec{u}_r/dt)$
so its $\vec{v}= (-sinθe^t)\vec{u}_r+(1+cosθ)e^t\vec{u}_θ.$
so acceleration vector.$\vec{a}=(d^2r/dt^2)\vec{u}_r+dr/dt(d\vec{u}_r/dt)+dr/dt(d\vec{u}_r/dt)+r(d^2\vec{u}_r/dt^2)$
.$\vec{a}=((-sinθ+-cosθθ).e^t)\vec{u}_r+2(-sinθe^t).(\vec{u}_θ.e^t)+(1+cosθ).(-\vec{u}_re^t+\vec{u}_θ)e^t$

Is this true ?

Yeah it is correct, good job.

 

[RyanH42]

Thanks

 

[Vatsal Sanjay]

This is not quite right. The tangential velocity is the component of the velocity vector tangent to the trajectory of the particle. This is not generally in the direction of uθ \theta . uθ \theta is the unit vector in the θ direction.

I must have used the word azimuthal. Thanks for pointing out.