Author Topic: Here's a question for you to try out. (Read 4501 times) Tweet Share

Lycan · « **on:** August 02, 2008, 02:50:45 pm »

I just finished a question that I thought was pretty cool so I thought I might give you guys a go.

Prove mathematically that if the speed of a moving particle moving along a curve is constant, the acceleration is perpendicular to the velocity.

The topic is vector calculus if you want a clue.

phagist_ · « **Reply #1 on:** August 02, 2008, 07:01:28 pm »

Constant velocity means acceleration is zero, not constant speed.

Collin Li · « **Reply #2 on:** August 02, 2008, 07:07:45 pm »

Constant velocity means that the acceleration is zero and that there is constant speed. (Ohh, okay, I get what you mean now, phagist_)

On the other hand, constant speed does not mean constant velocity.

Collin Li · « **Reply #3 on:** August 02, 2008, 07:09:17 pm »

Quote from: dcc on August 02, 2008, 04:07:42 pm

If a moving particle is moving with a constant speed, then the acceleration of the particle is zero.

Hence $\vec{v}.\vec{a} = \alpha \cdot \vec{0} = \vec{0}$ (where $\alpha$ is some constant), hence $\vec{v} \perp \vec{a}$

This result should hold true for any natural number of dimensions of movement.

This doesn't make sense. A dot product should give you a scalar quantity. How come you are getting the zero vector?

Collin Li · « **Reply #4 on:** August 02, 2008, 07:17:13 pm »

It still remains that a scalar times a vector is a vector, so that does not seem like a correct argument to me.

I would do it by proposing the velocity vector of the form: $(x,\, y)$ ,

and then using the fact that: $C = \sqrt{x^2 + y^2}$ , where $C$ is some constant,

hence we can find $y$ in terms of $x$ , and then you can find the dot product between the velocity vector and the acceleration vector.

Collin Li · « **Reply #5 on:** August 02, 2008, 07:29:03 pm »

Take: $\textbf{v} = x\,\textbf{i} + y\,\textbf{j}$

Since: $C = \sqrt{x^2 + y^2}$ , where $C$ is a constant,

$\implies y = \pm\sqrt{C^2 - x^2}$

$\implies \textbf{v} = x\,\textbf{i} \pm\sqrt{C^2 - x^2}\,\textbf{j}$

$\implies \textbf{a} = \frac{dx}{dt}\,\textbf{i} \pm \frac{d}{dt}\left(\sqrt{C^2 - x^2}\right)\,\textbf{j}$

$\implies \textbf{a} = \frac{dx}{dt}\,\textbf{i} \mp \frac{x}{\sqrt{C^2 - x^2}}\frac{dx}{dt}\,\textbf{j}$

Therefore:

$\textbf{v}\cdot\textbf{a} = x\frac{dx}{dt} - \sqrt{C^2 - x^2}\left[\frac{x}{\sqrt{C^2 - x^2}}\right]\frac{dx}{dt}$

(Note the plus/minus and minus/plus signs will always produce a negative sign)

$\tf\;\;\textbf{v}\cdot\textbf{a} = x\frac{dx}{dt} - x\frac{dx}{dt} = 0$

dcc · « **Reply #6 on:** August 02, 2008, 07:34:46 pm »

Im sorry, I misread the question as velocity, rather then speed.

Mao · « **Reply #7 on:** August 02, 2008, 07:36:54 pm »

Quote from: Lycan on August 02, 2008, 02:50:45 pm

I just finished a question that I thought was pretty cool so I thought I might give you guys a go.

Prove mathematically that if the speed of a moving particle moving along a curve is constant, the acceleration is perpendicular to the velocity.

The topic is vector calculus if you want a clue.

$|\vec{v}|=k$

$\vec{v}\cdot \vec{v}=|\vec{v}|^2=k^2$

$\frac{d}{dt}(\vec{v}\cdot \vec{v})=\frac{d}{dt}(k^2)$

    Assuming $\vec{v}$ is defined in $\mathbb{R}^3$

    $\vec{v}\cdot \vec{v}=x(t)\times x(t)+y(t)\times y(t)+z(t)\times z(t)$

    $\implies \frac{d}{dt}\left[x(t)\times x(t)+y(t)\times y(t)+z(t)\times z(t)\right]=x(t)\times \frac{dx}{dt}+\frac{dx}{dt}\times x(t)+y(t)\times \frac{dy}{dt}+\frac{dy}{dt}\times y(t) +z(t)\times \frac{dz}{dt}+\frac{dz}{dt}\times z(t)$ (product rule)

    $\implies \frac{d}{dt}(\vec{v}\cdot \vec{v})=\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)\cdot \left(x(t),y(t),z(t)\right) + \left(x(t),y(t),z(t)\right) \cdot\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)$

   (this can be shown for n-dimensions)

   $\implies \boxed{\frac{d}{dt}(\vec{v}\cdot \vec{v})=\frac{d\vec{v}}{dt}\cdot\vec{v}+\vec{v}\cdot \frac{d\vec{v}}{dt}}$ (basically, the product rule)

$\frac{d\vec{v}}{dt}\cdot\vec{v}+\vec{v}\cdot \frac{d\vec{v}}{dt}=0$

$2\vec{a}\cdot\vec{v}=0$

$\therefore \vec{a} \perp \vec{v}$ , given that neither $\vec{a}$ nor $\vec{v}$ are 0

hence, in general, where the magnitude of velocity is a constant, acceleration is perpendicular to the direction of motion

EDIT: and for dcc's sake, this holds true where velocity is defined and continuous (and in this case, its magnitude is also constant)

Collin Li · « **Reply #8 on:** August 02, 2008, 07:50:44 pm »

Moderator action: Unmerged by popular request.

enwiabe · « **Reply #9 on:** August 02, 2008, 07:51:29 pm »

coblin's request*

Collin Li · « **Reply #10 on:** August 02, 2008, 07:52:25 pm »

Ahmad, dcc and I versus Mao.

dcc · « **Reply #11 on:** August 02, 2008, 07:55:31 pm »

how about:

$\vec{v}(t) = \sum_{n = 1}^\infty \dfrac{\left(6 - 6 \cos n \pi \right) \sin \left(n \pi t \right)}{n \pi} \ \vec{i}$

Mao · « **Reply #12 on:** August 02, 2008, 07:56:51 pm »

Quote from: dcc on August 02, 2008, 07:55:31 pm

how about:

$\vec{v}(t) = \sum_{n = 1}^\infty \dfrac{\left(6 - 6 \cos n \pi \right) \sin \left(n \pi t \right)}{n \pi} \ \vec{i}$

is that constant magnitude?

dcc · « **Reply #13 on:** August 02, 2008, 08:00:32 pm »

The speed is constant (and sometimes undefined).

Mao · « **Reply #14 on:** August 02, 2008, 08:02:26 pm »

then for intervals where the velocity is continuous (and constant), acceleration will be perpendicular to the direction of motion.

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Author Topic: Here's a question for you to try out. (Read 4501 times) Tweet Share

Lycan

Here's a question for you to try out.

phagist_

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Collin Li

Here's a question for you to try out.

Collin Li

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Collin Li

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Collin Li

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dcc

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Mao

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Collin Li

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enwiabe

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Collin Li

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dcc

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