- #1

bolzano95

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^{2}/2 on the side. I also searched on the internet for a similar derivation, but there are none so simple.

Thanks for your help!

P.S There is a mistake in calculation in second line (textbook error).

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- Thread starter bolzano95
- Start date

- #1

bolzano95

- 89

- 7

Thanks for your help!

P.S There is a mistake in calculation in second line (textbook error).

- #2

Nugatory

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That's just the formula for displacement as a function of time when the acceleration is constant: ##s=\frac{1}{2}at^2##. You'll find many derivations if you google for "SUVAT", or you can work it out for yourself by starting with ##a=\frac{\mathrm{d}^2x}{\mathrm{d}t^2}## and integrating twice.I really don't understand how we get ##a\Delta{t}^2/2##

- #3

bolzano95

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But why are we looking at this motion from two different starting points? The first would be from a circle and the other one would be from an axis.

This gives me somehow some broken motion I don't quite understand. It becomes a contradiction with my understanding of uniform motion- that it is continuous and "smooth".

I'm trying to understand this without calculus involved :)

- #4

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For plane uniform circular motion the calculation is very simple. Let's put the motion in the ##xy## plane and the center of the circle in the origin. Then the position vector of the particle is

$$\vec{x}=r \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \\0 \end{pmatrix},$$

where ##r=\text{const}## is the radius of the circle and ##\omega=\text{const}## is the angular velocity.

From this you get the velocity and acceleration by simply taking the time derivatives,

$$\vec{v}=\dot{\vec{r}} = r \omega \begin{pmatrix} -\sin(\omega t) \\ \cos(\omega t) \\0 \end{pmatrix}$$

and

$$\vec{a}=\dot{\vec{v}}=-r \omega^2 \begin{pmatrix} \cos(\omega t) \\ \sin(\omega t) \\0 \end{pmatrix}=-\omega^2 \vec{x}.$$

If you also introduce the angular velocity as an (axial) vector, ##\vec{\omega}=(0,0,\omega)^{\text{T}}## then you can write the velocity as

$$\vec{v}=\vec{\omega} \times \vec{x}.$$

To get the magnitudes of velocity and momentum just take the square

$$\vec{v}^2=v^2=r^2 \omega^2 \; \Rightarrow \; v=\omega r$$

and

$$\vec{a}^2=r^2 \omega^4 \; \Rightarrow \; a=r \omega^2=\frac{v^2}{r}.$$

- #5

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It becomes a contradiction with my understanding of uniform motion- that it is continuous and "smooth".

I'm trying to understand this without calculus involved :)

You can't have uniform circular motion without calculus. Without calculus you can only do something like what is done in the diagram you posted.

- #6

vela

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Sorry, I should be more specific- why do we use aΔt^{2}/2 for the part from the point till intersection with vΔt? As I understand it, we don't look at this problem any more as a rotational motion, but as a straight motion of two points until they intersect.

But why are we looking at this motion from two different starting points? The first would be from a circle and the other one would be from an axis.

This gives me somehow some broken motion I don't quite understand. It becomes a contradiction with my understanding of uniform motion- that it is continuous and "smooth".

I remember when my high school physics teacher gave us this hand-waving argument about the direction of the centripetal acceleration, and it bugged me back then too because there's no reason to think that where the particle ends up on the circle, where the particle would have ended up if it followed the straight-line path, and the center of the circle are collinear. Indeed, it's easy to see that can't possibly be the case if ##\Delta t## is, for instance, one-fourth the period of rotation.

If ##\Delta t## is small enough, though, it's probably a good approximation along with taking the radial acceleration ##a## as constant. Then the correction to ##r## due to the acceleration would be just ##\frac 12 a(\Delta t)^2##. It feels to me, though, that some details are being glossed over, so I could understand why one might remain somewhat skeptical of the derived result.

You might find these explanations a bit more convincing: https://physics.stackexchange.com/q...ation-of-the-centripetal-acceleration-formula

- #7

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- #8

zoki85

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C=2r⋅π and A=r

I was curious how can we be absolutely sure that constant "π" is the same constant in both expression ?

I raised this question in front of a class and our teacher had a hard time trying to explain it.The problem was I wasn't satisfied with his naive explanation, asked additional questions, and he felt uncomfortable since I was best at math in his class. Happend looong time ago but I still remember of that episode

- #9

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When I first learned formulae for circumference and area of a circle in a primary school I wanted to know the following:

The formulae are C=2r⋅π and A=r^{2}⋅π

How can we be absolutely sure that constant "π" is the same constant in both expression.

I raised this question in front of a class and our teacher had a hard time trying to explain it.

The problem was I wasn't satisfied with his naive explanation, asked additional questions, and he felt uncomfortable since I was best at math in his class. Happend looong time ago but I still remember of that episode

It's a good question, but you should raise a new thread for this.

Another question to ask is how do we know all circles have the same ratio of ##C## to ##r## and ##A## to ##r^2##? All circles might look the same, but how can you prove or at least justify a common ratio for all circles?

- #10

zoki85

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Means of a projective geometry come to my mind ( doesn't necessarily involve calculus). But rigorous proof to identical π constant question for area and circumference ought to involve calculusAll circles might look the same, but how can you prove or at least justify a common ratio for all circles?

- #11

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Means of a projective geometry come to my mind ( doesn't necessarily involve calculus). But rigorous proof to identical π constant question for area and circumference ought to involve calculus

Calculus is best. But, you can look at a sequence of regular n-sided polygons and calculate the area and length of the perimeter of each. In this case, ##\pi## is defined as ##2\pi## radians in a circle.

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