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

• RyanH42
In summary, the conversation discusses the terms u## \vec{u}_r## and u## \vec{u}_θ## and their relation to the polar equation of a moving particle. The acceleration vector is also discussed in terms of unit vectors and their components. The correct expressions for the velocity and acceleration vectors are given.
RyanH42
< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

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

Last edited by a moderator:
RyanH42 said:
Actually I don't know the terms u⃗ r\vec{u}_r and u⃗ θ\vec{u}_θ
ur
is the vector along the radial velocity direction of the particle and u## \theta ## is the vector along the tangential velocity direction.

RyanH42 said:
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.

Vatsal Sanjay said:

ur
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

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 ?

RyanH42 said:
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 ?

RyanH42 said:
##\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

Chestermiller said:
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.

## 1. What do the ##\vec{u}_r## and ##\vec{u}_θ## vectors represent?

The ##\vec{u}_r## vector represents the radial direction, which is the direction pointing away from the origin. The ##\vec{u}_θ## vector represents the tangential direction, which is perpendicular to the radial direction and is tangent to the circle at a given point.

## 2. How are the ##\vec{u}_r## and ##\vec{u}_θ## vectors related?

The ##\vec{u}_r## and ##\vec{u}_θ## vectors are orthogonal, meaning they are perpendicular to each other. They also form a basis for polar coordinates, with ##\vec{u}_r## pointing in the radial direction and ##\vec{u}_θ## pointing in the tangential direction.

## 3. How are the ##\vec{u}_r## and ##\vec{u}_θ## vectors used in physics?

The ##\vec{u}_r## and ##\vec{u}_θ## vectors are commonly used in polar coordinate systems to describe the position and movement of objects. They are also used in mechanics, electromagnetism, and other areas of physics to calculate forces, velocities, and accelerations in circular motion.

## 4. Can the ##\vec{u}_r## and ##\vec{u}_θ## vectors be used in other coordinate systems?

Yes, the ##\vec{u}_r## and ##\vec{u}_θ## vectors can be used in other coordinate systems, such as cylindrical or spherical coordinates. In these systems, the ##\vec{u}_r## vector still points in the radial direction, while the ##\vec{u}_θ## vector points in a direction tangent to the surface of the coordinate system.

## 5. How can I calculate the components of the ##\vec{u}_r## and ##\vec{u}_θ## vectors?

The components of the ##\vec{u}_r## and ##\vec{u}_θ## vectors can be calculated using trigonometric functions. In polar coordinates, the ##\vec{u}_r## vector has a magnitude of 1 and an angle of 0 degrees, while the ##\vec{u}_θ## vector has a magnitude of 1 and an angle of 90 degrees. In other coordinate systems, the components can be calculated using similar methods based on the definitions of the vectors.

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