# I Rotational Motion - Understanding the Idea

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1. Feb 6, 2017

### Arman777

1- I am trying to understand the rotational motion and In most books theres $\vec θ$ unit vector which its, $\vec θ=(-sinθ,cosθ)$
I can see that If $\vec r$ unit vector is,$\vec r=(cosθ,sinθ)$ then ,

$\frac {d\vec r} {dt}=\vec θ$

Like calculating $\vec v=\frac {d\vec r} {dt}=R.\frac {dθ} {dt}.\vec θ$

I dont understand why we use this ? Just to make things easier ?

2-And we know that $\vec a=R∝\vec θ+Rw^2(-\vec r)$
here
$\vec {a_t}=R∝\vec θ$ and $\vec {a_r}=Rw^2(-\vec r)$

I know that If $a_t$ is zero then it becomes uniform circular motion.But In main equations there is an non-zero angular acceleration.So it means do we have to exert $\vec F=m(-R∝\vec θ)$ to create a uniform circular motion ?

Note:This is not a homework question so,I didnt ask there

2. Feb 6, 2017

### albertrichardf

In general yes. It simplifies rotational motion a lot. For instance, consider uniform circular motion. If you were to stick to $x$ and $y$ coordinates, they would be constantly changing and they become difficult to work with. They are best suited to translational motion because of this. If you use $r$ and $\theta$ coordinates you can immediately see that one derivative is zero, and that the other is simply the angular speed. The reason we use $\vec \theta$ is to give that speed a direction.

No we do need to apply such a force for uniform circular motion. The $\vec {a_t}$ has to be caused by a force. If there is no tangential force, then it is zero. The term appears in the formula because the formula is general, so it works for any value of $\vec {a_t}$, including zero. You obtain the value of $\vec {a_t}$ from the value of $\vec F_t$, not the other way round.

3. Feb 6, 2017

### Arman777

I understand now.Thanks

4. Feb 6, 2017

Welcome.