Square wheel and the speed of the point

[Poetria]

Homework Statement

Consider a bumpy road in which each hump has the shape
$y=\cosh(a)-\cosh(x)$ for $-a\leq x \leq a$
where $y'(a)=-1$ so $a=arcsinh(1)$

L=2
$(x_p(t), y_p(t)) = (t, \cosh(a)-\cosh(t))$
$(x_q(t), y_q(t)) = (t, \cosh(a))$

We place the square wheel onto the bumpy road so that the point C touches the point A when t=0. The wheel rolls along so that the point D lands on the point B. Hence, the size (the length of one side) of the square wheel is fixed at the value found previously.

Point A is at the top of the 0th hump, and
B is where the 0th hump and the 1st hump meet.
Q is the center of the wheel.
C is the middle of the side of the square which touches A at t=0
P is the point of contact of the wheel and the road.
At time t=0, the points A, C, and P coincide.
At all time t, the point P lies directly below the point Q.

If the wheel rotates at constant, unit speed, then the angle $\theta$ between P, Q and C satisfies $\frac {d\theta}{dt} = 1$
For $0\leq\theta\leq \frac \pi 4$ find the speed of Q as a function of t.

2. The attempt at a solution
I have taken the derivatives of x and y: $(x_q(t), y_q(t)) = (t, \cosh(a))$

$\sqrt{1+\sinh(t)^2}$

but this is wrong. :(

 

[kuruman]

https://my.vanderbilt.edu/stacyfonstad/files/2011/10/squareWheels.pdf

 

[Poetria]

https://my.vanderbilt.edu/stacyfonstad/files/2011/10/squareWheels.pdf

The height of the center of the wheel Q is a constant so its speed is also a constant and equals 1?

x=t
dx=1 ?

Many thanks. :)

 

[kuruman]

I am not sure what you are asking. The vertical height of point Q above the point of contact must be constant. The speed can be anything.

On edit: Yes, the speed is constant at one unit as stated. Sorry, I missed that.

 

[Poetria]

Oh OK. Many thanks. :) The question was about the speed of the point Q.

 

[Poetria]

Oh OK. Many thanks. :) The question was about the speed of the point Q.

Well, I have definitely screwed this up. There was a pitfall: dx/dtheta.

 

[kuruman]

If $(x_q(t), y_q(t)) = (t, \cosh(a))$ then
$\frac{d}{dt}(x_q(t), y_q(t)) = \frac{d}{dt}(t, \cosh(a))=(1, 0)$, no?

 

[Poetria]

If $(x_q(t), y_q(t)) = (t, \cosh(a))$ then
$\frac{d}{dt}(x_q(t), y_q(t)) = \frac{d}{dt}(t, \cosh(a))=(1, 0)$, no?

I thought like you but the correct solution is supposed to be sec(t)! I am trying to understand this. :( :(

 

[kuruman]

I suspect that the problem is asking to show that if $\frac{d\theta}{dt}$ is constant, the speed of Q is no longer constant. Let me think about this for a while.

 

[kuruman]

I think I got it. You have to work with the right triangle $QCP$ and observe that $QC=1$, $CP= \sinh(x)$ and $QP= \cosh(x)$. Also observe that $QC = QP \cos(\theta)$. This is also $1=\cos(\theta)\cosh(x)$. Take the time derivative (remember the chain rule), do some algebra and you should come up with the answer.

 

[Poetria]

Many thanks. :) I have to digest it. :)