# Can we write icap-jcap=pi/2 ? why or why not?

1. Oct 24, 2013

### AakashPandita

Can we write icap-jcap=pi/2 ? why or why not?

2. Oct 24, 2013

### mikeph

You can write whatever you want, but nobody will understand you if you don't explain what icap and jcap are!

3. Oct 24, 2013

### UltrafastPED

I think the OP is asking about the difference between the unit vectors in the i and j directions being 90 degrees.

Using that interpretation the answer is no; this difference (i - j) is a new vector that points along the line -45 degrees.

4. Oct 24, 2013

### Staff: Mentor

Interesting I took it to mean a programming statement in some language like Java:

Code (Text):

icap-jcap = pi/2;      // is an invalid construct
// as you can have an expression icap-jcap
// on the left of an assignment

but

Code (Text):

if ((icap-jcap)==(pi/2)) { System.out.println("icap-jcap == pi/2 is TRUE");

5. Oct 24, 2013

### AakashPandita

UltrafastPED is right.
I was trying to prove that radial acceleration of a body moving in circle is v^2/r.

For uniform speed i came up with this 2v^2(icap-jcap)/rpi as radial acceleration

If icap-jcap=pi/2, acc=v^2/r

Also i do not understand computer programs.

6. Oct 25, 2013

### mikeph

No, you can't do that. i and j are vectors, adding or subtracting them couldn't possibly give you a scalar.

7. Oct 25, 2013

### UltrafastPED

8. Oct 25, 2013

### nasu

There is also the problem that $$\hat{i}-\hat{j}$$ is a constant vector whereas the centripetal acceleration is a rotating vector. So your formula cannot be right.

9. Oct 25, 2013

### UltrafastPED

10. Oct 25, 2013

### Staff: Mentor

I see what you are trying to do. You are assuming that the particle is traveling in a circle in the clockwise direction, and at the top of the circle, the velocity is pointing in the i direction, while, earlier, at the left end of the circle, the velocity is pointing in the j direction. The time interval between these two locations is rπ/2 divided by v. You can't get the instantaneous acceleration this way because angular increment is too large. The best you can get is an approximation to the average acceleration over the interval. According to UltrafastPED's post #3, you get the correct direction for the average acceleration (at the half-way point), but not the correct magnitude. You get a coefficient of 2/π, and it should be unity (if you were trying to get the instantaneous magnitude at the half-way point). Still, considering the crudeness of the approximation, it's not too bad. You need to do the analysis in the limit as the angular change approaches zero.

11. Nov 2, 2013

### AakashPandita

okay okay now i think i understand.

12. Nov 2, 2013

### AakashPandita

thank you very much!

13. Nov 2, 2013

### AakashPandita

$$/frac{2v^2(î-{/jhat}}{rpi}$$

Last edited: Nov 2, 2013
14. Nov 2, 2013

### AakashPandita

2v^2(icap-jcap) / rπ

is the average acc. for 90 degree of circular motion.

is there a way i could obtain instantaneous acceleration from this expression?
how?

also how do i write it in latex?

Last edited: Nov 2, 2013
15. Nov 2, 2013

### Staff: Mentor

You would have to use an angle less that π/4, say Δθ, and, you could make use of the relationship for how the unit vector in the θ direction changes with θ:

$$\frac{d\vec{i_θ}}{dθ}=-\vec{i_r}$$

You would express the velocity vector as $\vec{V}=v\vec{i_θ}$, and then determine the rate of change of this vector with time.

16. Nov 2, 2013

### AakashPandita

yes i know that method. is there a proof .....like maybe we could find average acceleration for time tending to zero...?

17. Nov 2, 2013

### Staff: Mentor

Sure. Take Δθ larger, determine the vectorial velocity change over the angle Δθ, and then take the limit as Δθ approaches zero.

18. Nov 3, 2013

### AakashPandita

I get change in velocity:

$$\frac{( sinθ \hat{i} + ( cosθ - 1 ) \hat{j} ) v^2}{θr}$$

where θ is change in velocity and in radian and less than pi/2, speed is constant and body starts moving from
$$v\hat{j}$$

Last edited: Nov 3, 2013
19. Nov 3, 2013

### Staff: Mentor

Excellent. Now take the limit as θ approaches zero.

Chet

20. Nov 3, 2013

### AakashPandita

Am I going to get $$\frac{v^2}{r}$$ ?
because i am not getting it. I am getting something very messy.