Finding the velocity of $s$ using the chain rule

Question

Can I find the velocity of $s$ using the chain rule?

First I'll explain this diagram (sorry for the crappy quality, I had to draw it on Paint. Also im more of a maths person than an art person).

It's a rough sketch of a machine with an arm ($r$) which rotates anti-clockwise at velocity $v(P)$. The point where $r$ intersects with the circumference of the circle is $P$, where one side of the line $L$ is attached. The length of $L$ is constant. The other side of $L$ is attached to a horizontal surface (there doesnt have to be a surface, but the idea is that $s$ anly moves along this axis) which also passes thrugh the cnetre of the circle traced by the arm $r$ of the machine. $s$ moves with velocity $v(s)$.

I am trying to find $v(s)$ using the chain rule. The values of $v(P)$, r, L and $\emptyset$ are given.

My reasoning is as follows:

$$v(P)=\frac{dP}{dt}$$ where $t=$time. Same follows for $$v(s)=\frac{ds}{dt}$$

We know $\frac{dP}{dt}$ and we're trying to find $\frac{ds}{dt}$. This is a clear cut example of an application of the chain rule. Therefore I formulate the equation, using the chain rule, $$\frac{ds}{dt}=\frac{dP}{dt} \cdot \frac{ds}{dP}$$

Therefore I have to find $\frac{ds}{dP}$ in order to apply the chain rule and find the velocity of $s$. My problem is that I cant find an equation that relates $P$ and $s$.

I tried to find $S$ (distance between centre of circle and point $s$) by first finding the angle opposite to it and then using the cosine rule. If we call the angle opposite $\emptyset_S$ we can find it using the sine rule.

$$\emptyset_S=arcsin(\frac{L\cdot s}{\sin\emptyset})$$

Applying the cosine rule to find $S$,and thus having an equation relating both $s$ and $P$

$$s=\sqrt{r^2+L^2-2rL\cos\emptyset_S}$$

I would then have to derive this equation with respect to P, which im not sure how to do.

I dont know if im even on the right path trying to find $\frac{ds}{dP}$. Am I using the right equation? Is this how you find the velocity of $s$?

Any ideas, suggestions and corrections are welcome.

Answer 1

By the law of cosines:

$$L^2=r^2+s^2-2rs\cos\theta\tag{1}$$

Differentiating with respect to $t$:

$$0=2s\frac{ds}{dt}-2r\left[\frac{ds}{dt}\cos\theta-\sin \theta \frac{d\theta}{dt}\right]$$

Solving:

$$\frac{ds}{dt}=\frac{r\sin\theta}{r\cos\theta-s} \frac{d\theta}{dt}\tag{2}$$

Solving $(1)$:

$$s=r\cos\theta+ \sqrt{L^2-r^2\sin^2\theta}$$

and substituting into $(2)$ gives

$$\frac{ds}{dt}=-\frac{r\sin\theta}{\sqrt{L^2-r^2\sin^2\theta}} \frac{d\theta}{dt}$$

If the tangential velocity of $P$ is $v$ then $r\frac{d\theta}{dt}=v$ so that

$$\frac{ds}{dt}=-\frac{v\sin\theta}{\sqrt{L^2-r^2\sin^2\theta}}$$

For the case where tangential velocity $v$ is constant and $\theta(0)=0$ we would have $\theta(t)=vt/r$.

Answer 2

By the sine rule $$\frac{\sin{\theta}}{L}=\frac{\sin{\theta _s}}{r}=\frac{\sin{\theta _p}}{s}$$

$$\theta +\theta _s+ \theta _p=\pi$$

$$\implies \sin{\theta_p}=\sin{(\theta + \theta_s)}$$

Taking the time derivatives:

$$\frac{(\cos{\theta})\dot{\theta}}{L}=\frac{(\cos{\theta _s})\dot{\theta _s}}{r}=\frac{(\cos{\theta _p})\dot \theta _p}{s}-\frac{\sin{\theta _p}}{s^2}\dot s$$

$$\frac{\tan{\theta}}{\dot \theta}=\frac{\tan{\theta_s}}{\dot \theta _s}=\frac{s(\tan{\theta_p})}{\dot{\theta_p}s-\dot{s}\tan{\theta _p}}$$

$$\dot{s}=-r\sin{\theta}(\dot{\theta}+\dot{\theta _s})=-(r\sin{\theta})\dot{\theta _p}$$