Classical mechanics problem

**Mondreus** · November 27th 2010, 03:06 PM

I need to find the slope $\frac{dy}{dx}$ at the point P. The only thing I know is that:

$y = f(x)$
$\frac{dx}{dt} = v_0$ (constant)
The acceleration at P is: $\vec{a} = \frac{a}{\sqrt{2}}(\hat{t}+\hat{n})$ (a constant and > 0, natural coordinates)

**CaptainBlack** · November 27th 2010, 10:47 PM

Originally Posted by Mondreus

I need to find the slope $\frac{dy}{dx}$ at the point P. The only thing I know is that:

$y = f(x)$
$\frac{dx}{dt} = v_0$ (constant)
The acceleration at P is: $\vec{a} = \frac{a}{\sqrt{2}}(\hat{t}+\hat{n})$ (a constant and > 0, natural coordinates)

Post the actual question (or try defining all the symbols), that is incomprehensible.

CB

**Mondreus** · November 27th 2010, 11:08 PM

A ball moves along the curve $y=f(x)$ with the constant velocity $\dot{x}=v_0$ . At a certain position P, the ball has the acceleration $\vec{a} = \frac{a}{\sqrt{2}}(\hat{t}+\hat{n})$ , $a > 0$

What is the slope $\frac{dy}{dx}$ at P.

The acceleration is given in normal-tangential coordinates, i.e. $\vec{a} = \ddot{s}\hat{t}+\frac{\dot{s}^2}{\rho}\hat{n}$

**Ackbeet** · November 29th 2010, 06:05 AM

Well, isn't it true that

$\dfrac{\dot{y}}{\dot{x}}=\dfrac{dy/dt}{dx/dt}=\dfrac{dy}{dt}\,\dfrac{dt}{dx}=\dfrac{dy}{dx}?$

Since you're given $\dot{x},$ it remains to find $\dot{y}.$ How do you suppose you could do that?

**Mondreus** · November 29th 2010, 10:37 AM

It's probably really easy, but I can't figure it out. It's actually a two-part question where I also need to find the radius of curvature, but that one I can solve once I know dy/dx by using the arc length and identifying $\rho$ with $\dot{s}$ in the acceleration that we know.

$ds = \sqrt{1+\left(dy/dx \right) ^2}dx \implies \dot{s}=\dot{x}\sqrt{1+\left(dy/dx \right) ^2}$

**Ackbeet** · November 29th 2010, 11:13 AM

In your post # 3, what is $\rho?$ Is it known? Is $f(x)$ known? If so, what is it?

**Mondreus** · November 29th 2010, 12:43 PM

$\rho$ is the radius of curvature, and it's unknown (I can find it once I know dy/dx). $f(x)$ is also unknown.

**Ackbeet** · November 29th 2010, 05:41 PM

So, what is the representation of the velocity vector in terms of tangential and normal components? How does taking the derivative of that compare with the acceleration in tangential and normal components? Could you infer anything from that?

**Mondreus** · November 29th 2010, 05:59 PM

$\vec{v} = \dot{s}\hat{t}$ where $\hat{t}$ is defined as $\frac{d\vec{r}}{ds}$ and where $s$ is the arc length and $\vec{r}$ is the distance vector from the origin.

$\frac{\dot{s}}{\rho} = \frac{a}{\sqrt{2}}$ is given, but $\rho$ is unknown.

The answer is $\frac{dy}{dx}\pm 1$ , but I don't know how to get there.

**Ackbeet** · November 30th 2010, 02:42 AM

So what happens when you differentiate both sides of

$\vec{v}=\dot{s}\,\hat{t}$

w.r.t. time?

**Mondreus** · November 30th 2010, 06:52 AM

You get the acceleration I posted above: $\vec{a} = \ddot{s}\hat{t}+\frac{\dot{s}^2}{\rho}\hat{n}$

$\rho$ is defined as $\frac{1}{\left|\frac{d\hat{t}}{ds} \right|}$

**Ackbeet** · November 30th 2010, 07:24 AM

So what happens when you compare the two accelerations? Say, just the tangential component?

**Mondreus** · November 30th 2010, 10:24 AM

$\ddot{s} = \frac{a}{\sqrt{2}}$

Don't know how that helps me since the acceleration is only given for a single point, so I can't integrate it to get the velocity.

Comparing the normal components gets us $\frac{\dot{s}^2}{\rho} = \frac{a}{\sqrt{2}}$

**Ackbeet** · November 30th 2010, 10:35 AM

I can't integrate it to get the velocity.

Perhaps not exactly. But you could get it correct to first order, perhaps. That might be good enough.

Question: how do you interpret the phrase "constant velocity $\dot{x}=v_{0}$ "? Do you think that $\dot{x}=v_{0}$ for all time? Do you also think that initially, the ball is moving only in the x direction with speed $v_{0}?$

**Mondreus** · November 30th 2010, 10:42 AM

It is moving with constant velocity $\dot{x}=v_0$ all the time.

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