This derivation determines the radial part of the total acceleration. The tangent part is determined with a tangent part of an external force.

That derivation shows that there is only radial acceleration.

 

If there was another component the derivation should not be correct

=> [math]\hat{\mathbf{r}}[/math].a = [math]\frac{-v^2}{r}[/math]

 

multiply by :[math]\hat{\mathbf{r}}[/math] both sides of the equation and:

 

a= -[math]\hat{\mathbf{r}}\frac{v^2}{r}[/math]

Let's use that logic on the velocity vector.

 

[math]\hat r \cdot \vec v = 0[/math]

 

Multiply by [math]\hat r[/math] on both sides and:

 

[math]\vec v = 0[/math]

The velocity vector is identically zero!

 

Which of course is wrong, and it is wrong for exactly the same reason your derivation was wrong.

Let's use that logic on the velocity vector.

 

[math]\hat r \cdot \vec v = 0[/math]

 

Multiply by [math]\hat r[/math] on both sides and:

 

[math]\vec v = 0[/math]

The velocity vector is identically zero!

 

Which of course is wrong, and it is wrong for exactly the same reason your derivation was wrong.

 

[math]\hat{\mathbf{r}}[/math] and v are rerpendicular to each other that's why you get that result .

 

But r and a are not perpendicular to each other.

 

An analogical example is not enough you have to pinpoint the mistake if there is one

What you did correctly, triclino, was to calculate the component of the acceleration vector in the direction of the radial vector. What you did wrong was to assume that this is equal to the acceleration vector.

 

In particular, you calculated [math](\hat r \cdot \vec a)\hat r[/math], and this is equal to the acceleration vector only in the case that the acceleration vector is parallel to the radial vector. The transverse component of the acceleration vector is not zero in the case of non-uniform circular motion.

What you did correctly, triclino, was to calculate the component of the acceleration vector in the direction of the radial vector. What you did wrong was to assume that this is equal to the acceleration vector.

 

In particular, you calculated [math](\hat r \cdot \vec a)\hat r[/math], and this is equal to the acceleration vector only in the case that the acceleration vector is parallel to the radial vector. The transverse component of the acceleration vector is not zero in the case of non-uniform circular motion.

 

I assumed nothing. All i did is certain mathematical operations involving vector Analysis.

 

The only assumption in the whole procedure is that :r andv

are perpendicular.

 

The final result was dictated by the rules of vector analysis.

 

Unless the rules i used are not correct.

Unless the rules i used are not correct.

You missed one possibility: That you misinterpreted/misapplied them, and that is what happened. What you calculated was the orthogonal projection of the acceleration vector a onto the line defined by the radial vector r. There is nothing wrong with that; in fact that is a frequently used operation. Your mistake was assuming that this orthogonal projection is equal to the acceleration vector. That is not the case.

 

As a side exercise, I suggest you investigate [math]d/dt\left(\vec r \times \vec v\right)[/math] , and also the vector triple product, [math]\hat r\times(\vec a \times \hat r)[/math], in the case of non-uniform circular motion.

 

You might also want to re-reread post #20.

You missed one possibility: That you misinterpreted/misapplied them, and that is what happened. What you calculated was the orthogonal projection of the acceleration vector a onto the line defined by the radial vector r. There is nothing wrong with that; in fact that is a frequently used operation. Your mistake was assuming that this orthogonal projection is equal to the acceleration vector. That is not the case.

 

As a side exercise, I suggest you investigate [math]d/dt\left(\vec r \times \vec v\right)[/math] , and also the vector triple product, [math]\hat r\times(\vec a \times \hat r)[/math], in the case of non-uniform circular motion.

 

You might also want to re-reread post #20.

.

 

 

 

The rules i used are the following:

 

1) if A and B are vectors then :

 

[math]\frac{d(A.B)}{dt}=\frac{dA}{dt}.B + A.\frac{dB}{dt}[/math]

 

2) A.A = [math]A^2[/math]

 

3) k(A.B) = (kA).B ,where k is a real No

 

4) If we add the same No to both sides of an equation the equation does change

 

5) if we multiply with the same No both sides of the equation the equation does not change .

 

 

6) the dot product of unit vectors is 1.

 

Now where is the misapplication??

Your mistake is treating vectors as numbers. While numbers do form a vector space ([math]\mathbb R^1[/math]), vector spaces have different rules than numbers. In particular, you treated the inner product and scalar multiplication as interchangeable operations. They are not. Given vectors [math]\vec a[/math] and [math]\vec b[/math], [math](\vec a \cdot \vec b)\vec b \ne \vec a (\vec b \cdot \vec b)[/math] in general. That is exactly what you did, and that is invalid.

if k is scalar then only no 3 is true.

=D H;556451]Your mistake is treating vectors as numbers. While numbers do form a vector space ([math]\mathbb R^1[/math]), vector spaces have different rules than numbers.

 

 

 

 

The rules i just quoted are those of vector spaces and some of those rules coincide with those of a field .

 

For example those concerning the predicate of equality.

 

Commutativity.

 

Existence of identity element.....e.t.c........e.t.c

 

 

 

 

 

In particular, you treated the inner product and scalar multiplication as interchangeable operations.

 

 

 

You mean k(A.B) = (kA).B is not correct ,k being a real.

 

 

 

 

 

 

 

They are not. Given vectors [math]\vec a[/math] and [math]\vec b[/math], [math](\vec a \cdot \vec b)\vec b \ne \vec a (\vec b \cdot \vec b)[/math] in general. That is exactly what you did, and that is invalid.

.

 

 

 

No where in my proof i did that mistake

.

Yes, you did. Your continued trolling is starting to get old.

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