# Acceleration Vector and Motion of Particle

1. Jun 18, 2015

### RyanH42

1. The problem statement, all variables and given/known data
Show that the acceleration of a particle is always normal to its path of motion if and only if its speed is constant.

2. Relevant equations
1-$\vec{a}=d^2s/dt^2\vec{T}+κ(ds/dt)^2\vec{N}$
2- $\vec{T}=(d\vec{R}/dt)/(\|d\vec{R}/dt\|)$
3-$\vec{v}=d\vec{R}/dt$
4-$\|\vec{v}\|=\|ds/dt\|$

3. The attempt at a solution
If speed is constant then $d\vec{v}/dt=0$.I have to use first equation to make conclusion.Now,First equation obtained by differantiation of $v=(ds/dt)\vec{T}$.Now If we differantiate it respect to t we get this
$\vec{a}=d^2s/dt^2\vec{T}+(ds/dt)(d\vec{T}/dt)$, since $d\vec{T}/dt=0$, why this equation is true,cause of equation 2 and 3.The answer says this ; $\vec{a}=0\vec{T}+κ(ds/dt)^2\vec{N}$. Why $\vec{T}$-componenet is zero cause if $\|\vec{v}\|=\|ds/dt\|$ then $\|(d\vec{v}/dt)\|=0$ cause $\|ds/dt\|$ is constant .So I am saying both sides has to be zero.But here not Why ?I am missing something but I cant see it.Maybe I made a wrong math.

Quote from answer: From .$\vec{a}=0\vec{T}+κ(ds/dt)^2\vec{N}$ $\vec{a}$ is a scalar multiple of $\vec{N}$ so that the acceleration is normal to the path in this case.
Conversely, if we now assume that the acceleration is normal to the path then $\vec{T}$-componenet of acceleration must be $0$. But the $\vec{T}$-component is $d^2s/dt^2$,since is $d^2s/dt^2=0$,$ds/dt$ is constant, and the assertion is proved.

2. Jun 18, 2015

### Nathanael

This is not true. Even with a constant speed, if the curve is anything but a straight line then the direction of velocity (and thus $\vec{v}$) will change.

This equation pretty much says it all. If the speed is constant then what is $d^2s/dt^2$?

3. Jun 18, 2015

### RyanH42

I understand the idea of $d^2s/dt^2=0$.I didnt understand why $d\vec{T}/dt$ is not zero ?

4. Jun 18, 2015

### Nathanael

$d\vec{T}/dt$ will be zero only if the path is straight.

$\vec{T}$ is a unit vector so it can't change in size, but it can point in different directions. If the path curves then $d\vec{T}/dt$ will not be zero because $\vec{T}$ will be changing directions.

5. Jun 18, 2015

### Nathanael

This is not true: $\|(d\vec{v}/dt)\|=0$,

It should be this: $\frac{d}{dt}(\|\vec{v}\|)=0$

6. Jun 18, 2015

### RyanH42

Ok,I understand.I assumed $d^2R/dt^2$ will be zero.Why cause I assumed $dv/dt$ will be zero.I didnt understand something.You said $dv/dt$ cant be zero.So $v$ is constant it means theres a no $t$ in there I mean the v can be something like this $\vec{v}=1\vec{i}+23\vec{j}$..This is constant $\vec{v}$.Can you give me an example of constant $\vec{v}$ which I did Then I will be understand better the idea.

7. Jun 18, 2015

### Nathanael

$d\vec v/dt$ can be zero, but not always.

That is an example where $d\vec v/dt$ is not zero. That means it must be a straight-line path.
But if the path is not straight, then $d\vec v/dt$ will not be zero (even if the speed is constant).

You want an example of constant speed with non-zero acceleration? Anything traveling at a constant speed in a curved path is an example.
Take circular motion for example:

$\vec{v}=cos(t)\hat i+sin(t)\hat j$
Speed $=\|\vec v\|=\sqrt{cos(t)^2+sin(t)^2}=1=$ constant
$\vec a=\frac{d\vec v}{dt}=-sin(t)\hat i+cos(t)\hat j=$ not constant

Since the speed is constant, the acceleration must be normal to the velocity. We can check this: $\vec a \cdot \vec v = -sin(t)cos(t) + cos(t)sin(t)=0$

8. Jun 18, 2015

### RyanH42

I have just finished thee high school and I am practicing myself to get ready to university.Thats why I am getting trouble to understand the idea.I understand the idea thanks for help.

9. Jun 18, 2015

### Nathanael

It is better to keep questioning than to pretend to understand. Good luck at university you will do well.

10. Jun 18, 2015

### RyanH42

Thanks

11. Jun 18, 2015

### haruspex

The easiest way is to consider $\|\vec v\|^2=\vec v.\vec v$. What is the derivative of the second of those?

12. Jun 18, 2015

### RyanH42

$2(d\vec{v}/dt)\vec{v}=d\|\vec{v}\|^2/dt$

13. Jun 18, 2015

### haruspex

Right, or as I would write it, $2\vec v.\dot{\vec v} = \frac d{dt}\|v\|^2$.
If you are told that the speed is constant, what does that equation tell you about the relationship between $\vec v$ and $\dot {\vec v}$?
Is that connection "if and only if"?

14. Jun 19, 2015

### RyanH42

The lenght of v is constant so derivative of it will be zero then the dot product of v and a will be zero.It means v and a are perpandicular each other

15. Jun 19, 2015

### haruspex

Right. And the converse?

16. Jun 19, 2015

### RyanH42

I did not understand converse.Is there something to add to this info

17. Jun 19, 2015

### haruspex

In post #14, you used the equation in post #12 to show that if the speed is constant then velocity and acceleration are perpendicular. To complete the task, you need to show that if velocity and acceleration are perpendicular then the speed is constant. Can you do that using the same equation?

18. Jun 19, 2015

### RyanH42

$\vec{v}.\vec{a}=0$ then İf the magnitude of v is constant the magnitude of a will be 0 so I mean If and only If the magnitude of v will be constant to make a and v perpandicular.Or make dot product zero.Here the angle between a and v is what If you say 90 degrees (I know thts 90 cause they are perpandıcular) then theres no need constant v.Cause the angle can be 90 and cos90 is zero.Here I am confused constant v is not necessery to make a and c perpandicular.If the angle between them is 90 then they will also perpandicular.