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

[AakashPandita]

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

 

[mikeph]

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

 

[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.

 

[jedishrfu]

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

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

but

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

 

[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.

 

[mikeph]

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

 

[UltrafastPED]

Show your work ...

 

[nasu]

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.

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.

 

[UltrafastPED]

Or study this paper: http://msx4.pha.jhu.edu/henryDir/pubsPDFDir/circularMotion.pdf

 

[Chestermiller]

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.

 

[AakashPandita]

okay okay now i think i understand.

 

[AakashPandita]

thank you very much!

 

[AakashPandita]

/frac{2v^2(î-{/jhat}}{rpi}

 

[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?

 

[Chestermiller]

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?

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.

 

[AakashPandita]

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

 

[Chestermiller]

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

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

 

[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}

 

[Chestermiller]

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}

Excellent. Now take the limit as θ approaches zero.

Chet

 

[AakashPandita]

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

 

[AakashPandita]

this is a vector . how do we differentiate a vector?

 

[Chestermiller]

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}

Your sign looks wrong on the sine term, and you should really have Δθ instead of θ. But, otherwise it's OK. So you should have:
\vec{a}=\frac{( -sinΔθ \hat{i} + ( cosΔθ - 1 ) \hat{j} ) v^2}{Δθr}

You basically did the vector differentiation correctly. Now, take the limit as Δθ approaches zero.

Lim(sin(Δθ)/Δθ) as Δθ approaches zero is 1
Lim((cos(Δθ)-1)/Δθ) as Δθ approaches zero is 0

So:
\vec{a}=- \hat{i}\frac{v^2}{r}

This is the acceleration vector, and it's pointing toward the origin (as it should).

 

[AakashPandita]

i don't understand how sinΔθ icap is negative.

 

[AakashPandita]

if the angle is less than 90 degrees shouldn't sintheta and costheta both be positive?

 

[Chestermiller]

Draw a diagram, and you'll see why.

chet

 

[AakashPandita]

clearly sinθ is positive?

 

[Chestermiller]

clearly sinθ is positive?

I'm sorry. I was thinking of it going counter clockwise starting at (r,0). The way you have it drawn is consistent with your equation. And, as your result indicates, it should be a +i unit vector (which points at the origin).

 

[AakashPandita]

Thanks a lot!