Finding unknown quadratic equations

[madgab89]

Homework Statement

In the attached diagram, the displacement of a follower (y) changes as the angle of the cam changes ($$\vartheta$$). The follwer has a smooth velocity over the entire path. Therefore, the velocity must match on both sides of the four points noted on the graph.

If we assume all curves are parabolas, what would their equations be and what is the value of $$\vartheta_{1}$$

Homework Equations

I came up with :
y=a₁x²
y^'=2a₁x

and

y=a₂ (x+2$$\pi$$/3)²
y^'=2a₂(x-2$$\pi$$/3)

The Attempt at a Solution

Now I don't know where to go from here, how am I supposed to solve for theta? I have too many unknowns. I know y' is supposed to be the same throughout the graph so would setting the two y' equations equal accomplish anything?

 

[n00bhaus3r]

I think I did a similar problem with a rollercoaster. Basically, 1 and 2 must intersect and 3 and 4 must intersect at their respective points. The value of their first derivatives must be the same because it must be continuous, and the value of their second derivatives must be the same for them to be "smooth." Hope that helps.

 

[madgab89]

Yeah that's pretty much what I figured but I don't know what I'm supposed to do with all these unknowns... I really just don't know where to start.

 

[n00bhaus3r]

Yes. It becomes a huge systems of equations (that are probably best solved by matrices).

 

[HallsofIvy]

I started a couple of times to respond to this, then stopped because it looked to complex. Once I sat down and wrote out what we know, it isn't that complex.

Since the first parabola has a horizontal tangent at (0,0), it can be written $y= a_1x^2$. We also know that at $x= \theta_1$, y= 3/4 so $a_1\theta_1^2= 3/4$.

The second parabola has a horizontal tangent at $(2\pi/3, 2)$ so it can be written as $y= a_2(x- 2\pi/3)+ 2$. At $x= \theta_1$, y is still 3/4 so $a_2(\theta- 2\pi/3)+ 2= 3/4$. Finally, the two derivatives must be equal at $\theta_1$: $2a_1\theta_1= 2a_2(\theta_1- 2\pi/3)$.

That gives 3 equations to solve for $a_1$, $a_2$ and $\theta_1$:
$a_1\theta_1^2= 3/4$
$a_2(\theta- 2\pi/3)+ 2= 3/4$
$2a_1\theta_1= 2a_2(\theta_1- 2\pi/3)$

A little thought about "symmetry" should show how to get the coefficents for the last two parabolas from those without having to solve any more equations.

 

[madgab89]

Oh my goodness! If I could hug you over the internet I would :P Thank you so much! I can't believe I didn't realize how simply this could be broken down...i think I was just overthinking it.

 

[madgab89]

Which variable would be the easiest to start solving for?

 

[grint]

I started a couple of times to respond to this, then stopped because it looked to complex. Once I sat down and wrote out what we know, it isn't that complex.

Since the first parabola has a horizontal tangent at (0,0), it can be written $y= a_1x^2$. We also know that at $x= \theta_1$, y= 3/4 so $a_1\theta_1^2= 3/4$.

The second parabola has a horizontal tangent at $(2\pi/3, 2)$ so it can be written as $y= a_2(x- 2\pi/3)+ 2$. At $x= \theta_1$, y is still 3/4 so $a_2(\theta- 2\pi/3)+ 2= 3/4$. Finally, the two derivatives must be equal at $\theta_1$: $2a_1\theta_1= 2a_2(\theta_1- 2\pi/3)$.

That gives 3 equations to solve for $a_1$, $a_2$ and $\theta_1$:
$a_1\theta_1^2= 3/4$
$a_2(\theta- 2\pi/3)+ 2= 3/4$
$2a_1\theta_1= 2a_2(\theta_1- 2\pi/3)$

A little thought about "symmetry" should show how to get the coefficents for the last two parabolas from those without having to solve any more equations.

I believe one of the equations is wrong, you forgot a square.
$a_2(\theta- 2\pi/3)+ 2= 3/4$ the part in the parenthesis should be squared