Question regarding parametric equations

Question

A circle of radius 2 rolls along a flat surface at a constant rate, as shown below. A point which has a distance of 3 from the center of the circle is fixed with respect to the circle.

Let $A$ be the midpoint of the arc, and let $B$ be a point where the path of the point intersects the surface. Let $v_A$ and $v_B$ be the speeds of the point at $A$ and $B,$ respectively. Compute $\frac{v_A}{v_B}.$

How do I set up a parameterization and then find the speeds?

Answer 1

The center of the circle follows this path.

$x = 2\theta, y = 2$

If the radius was something other than 2, you would plug that value.

A point moving in a circular path of radius 3.

$x = 3\cos \theta, y = 3\sin \theta$

But this is a counter-clockwise rotation, and first set of equations reflect a clockwise rotation. And lets change these such that the minimum y values is associated with $t = 0$

$x = -3\sin \theta, y = -3\cos \theta$

And we can just add these two components of motion together.

$x = 2\theta - 3\sin \theta, y = 2 - 3\cos\theta$

To find velocity.

$v_x = \frac {dx}{d\theta} = 2 - 3\cos\theta\\ v_y = \frac {dy}{d\theta} = 3\sin\theta$

When $\theta = 0, (v_x,v_y) = (-1,0)$

speed $\|v\| = 1$

To find $A$ we will need to find a second value of $\theta$ where

$x = 2\theta_2 - 3\sin \theta_2 = 0$

You will need numerical methods.

Once you have $\theta_2$

finding $v(\theta_2)$ is fairly simple.

Answer 2

Starting from @Doug M's answer, you need to solve for $t=\theta_2$ the equation $$x=2t-3\sin(t)=0$$ which requires numerical method (not a big deal).

We can make approximations of the solution considering the function $$f(t)=2t-3\sin(t)$$ for which $$f'(t)=2-3\cos(t) \qquad \text{and} \qquad f''(t)=3\sin(t)$$ The first derivative cancels at $$t_*=\cos ^{-1}\left(\frac{2}{3}\right)\implies f(t_*)<0 \qquad\text{and} \qquad f''(t_*)>0$$ Make a Taylor series around $t_*$ to get $$f(t)=\left(2 t_*-\sqrt{5}\right)+\frac{1}{2} \sqrt{5} \left(t-t_*\right)^2+O\left(\left(t-t_*\right)^3\right)$$ Then $$t_0=t_*+\sqrt{\frac{2}{5}} \sqrt{5-2 \sqrt{5} t_*}\approx 1.54495$$ while the exact solution, obtained using Newton method, is $1.49578$.

Taking into account that this result is close to $\frac \pi 2$, let us make a Taylor expansion of $f(t)$ around this new point $$f(t)=(\pi -3)+2 \left(t-\frac{\pi }{2}\right)+\frac{3}{2} \left(t-\frac{\pi }{2}\right)^2-\frac{1}{8} \left(t-\frac{\pi }{2}\right)^4+\frac{1}{240} \left(t-\frac{\pi }{2}\right)^6+O\left(\left(t-\frac{\pi }{2}\right)^8\right)$$ and use series reversion to get $$t=\frac{\pi }{2}+x-\frac{3 x^2}{4}+\frac{9 x^3}{8}-\frac{131 x^4}{64}+\frac{531 x^5}{128}-\frac{69001 x^6}{7680}+\frac{104229 x^7}{5120}+O\left(x^{8}\right)$$ where $x=\frac{3-\pi }{2}$. This leads to $t=1.49578160$ while the exact solution is $1.49578157$.

Another thing we could do is to build the $[1,n]$ Padé approximant of $f(t)$ around $t=\frac \pi 2$ and get explicit approximations $$\left( \begin{array}{ccc} 0 & \frac 3 2 &1.500000000 \\ 1 & \frac{-24-9 \pi +3 \pi ^2}{-34+6 \pi } & 1.496030120 \\ 2 & \frac{-408+24 \pi ^2}{-416+96 \pi } & 1.495794286 \\ 3 & \frac{9984-72 \pi -1032 \pi ^2+72 \pi ^3+8 \pi ^4}{11120-3600 \pi +144 \pi ^2+16 \pi ^3} & 1.495782202 \\ 4 & \frac{266880-29952 \pi -26880 \pi ^2+3840 \pi ^3+128 \pi ^4}{290816-118272 \pi +9216 \pi ^2+512 \pi ^3} & 1.495781599 \end{array} \right)$$

Now, using Newton method, the successive iterates would be $$\left( \begin{array}{cc} n & t_n \\ 0 & 1.570796327 \\ 1 & 1.500000000 \\ 2 & 1.495796460 \\ 3 & 1.495781568 \end{array} \right)$$

Answer 3

We need to find a parametric representation of the center of the circle. Let $t$ represent time. We place the center of the circle at $(t,2).$

The circumference of the circle is $2 \pi (2) = 4 \pi,$ so the circle makes a full revolution when $t = 4 \pi.$ Then relative to the center of the circle, the point which has a distance of 3 from the center of the circle has an $x$-coordinate of $-3 \sin \frac{t}{2},$ and $y$-coordinate of $-3 \cos \frac{t}{2},$ so the path of the point is parameterized by $$\left( t - 3 \sin \frac{t}{2}, 2 - 3 \cos \frac{t}{2} \right).$$ Then $\frac{d}{dt} \left( t - 3 \sin \frac{t}{2} \right) = 1 - \frac{3}{2} \cos \frac{t}{2},$ and $\frac{d}{dt} \left( 2 - 3 \cos \frac{t}{2} \right) = \frac{3}{2} \sin \frac{t}{2},$ so the speed of the point is \begin{align*} \sqrt{\left( 1 - \frac{3}{2} \cos \frac{t}{2} \right)^2 + \left( \frac{3}{2} \sin \frac{t}{2} \right)^2} &= \sqrt{1 - 3 \cos \frac{t}{2} + \frac{9}{4} \cos^2 \frac{t}{2} + \frac{9}{4} \sin^2 \frac{t}{2}} \\ &= \sqrt{1 - 3 \cos \frac{t}{2} + \frac{9}{4}} \\ &= \sqrt{\frac{13}{4} - 3 \cos \frac{t}{2}}. \end{align*} For point $A,$ we can take $t = 2 \pi.$ Then $$v_A = \sqrt{\frac{13}{4} - 3 \cos \frac{2 \pi}{2}} = \frac{5}{2}.$$ For point $B,$ the $y$-coordinate is 0. Then $2 = 3 \cos \frac{t}{2},$ so $\cos \frac{t}{2} = \frac{2}{3}.$ Then $$v_B = \sqrt{\frac{13}{4} - 3 \cdot \frac{2}{3}} = \frac{\sqrt{5}}{2}.$$Hence, $\frac{v_A}{v_B} = \frac{5/2}{\sqrt{5}/2} = \boxed{\sqrt{5}}.$

Answer 4

The description of the problem is ambiguous (what, for instance, is the arc of which $\ A\ $ is the mid-point), and my interpretation seems to be different from those adopted by the other two answers. I'm assuming the point $\ A\ $ is at one of the points where the arc is at its maximum height of $\ 5\ $ units above the surface on which the circle is rolling, as depicted in the diagram below. I'll take the $\ x$-axis to be along this surface in the same direction as the circle is rolling, the origin $\ O\ $ to be the point on the surface immediately below $\ A\ $, and the $\ y$-axis to be along the line from $\ O\ $ to $\ A\ $. After the circle has rotated through an angle of $\ \theta\ $ radians, its centre will have moved $ \ 2\theta \ $ units, and the coordinates of the moving point will be \begin{align} x&=2\theta+3\sin\theta\\ y&=2+3\cos\theta \ , \end{align} as illustrated in the diagram below. The components of its velocity will be \begin{align} \dot{x}&=\dot{\theta}\,(2+3\cos\theta)\\ \dot{y}&=-3\dot{\theta}\sin\theta \ , \end{align}and its speed will be $$ v=\sqrt{\dot{x}^2+\dot{y}^2}=\dot{\theta}\sqrt{\mathstrut4+12\cos\theta+9}\ . $$ At $\ A\ $ we have $\ \theta=0\ $, so $\ v_A=5\dot{\theta}(0)\ $, and at any point $\ B\ $ where the moving point is crossing the surface, we have $\ y=0\ $, or $\ \theta=-\frac{2}{3}\ $, so $\ v_B=\sqrt{5}\dot{\theta}\big(t_B\big)\ $. If the motion is uniform, then $\ \dot{\theta}\ $ is constant, and $$ \frac{v_A}{v_B}=\sqrt{5}\ . $$