Hockey pucks and parameters

Question

There is one hockey puck with a diameter of $3$ inches. The puck is spinning around its center at a speed of $3$ counterclockwise rotations per second. At the center, the puck is traveling at a speed of $24$ inches per second at an angle of $45^\circ$ to the positive $x$-axis.

(a) At time $t=0$, the center of the puck is at the origin. Find the location of the center of the puck at time $t$? (Note that time $t$ is measured in seconds.)

(b) A point on the outer edge of the puck begins at the point $\left(0,\frac32\right)$. Find its location at time $t$?

I am not sure how to find part (a), but this is what I have for part b).

The puck has diameter of $3$ inches, a radius of $\frac{3}{2}$.

I used the formula where that in order to find a point on the circle's circumference, given a circle with center $(a,b)$ and radius $r$,

$x(t) = r\cos (t) + a$ $y(t) = r \sin (t) + b$

the circle's center is at $(0,0)$ because of part a) statement.

Therefore $x(t) = r \cos (t)$ and $y(t) = r \sin (t)$. I plugged in $x(t) = 0$ and $y(t) = \frac{3}{2}$ because of what was given in part b). From there I found that $t = \frac{\pi}{2}$ and that $r = \frac{3}{2}$,

so now the formula is $x(t) = \frac{3}{2} \cos (t)$ and $y(t) = \frac{3}{2} \sin (t)$. But now I am stuck. What do I do next for part b), and how do I solve part a)?

Answer 1

If you don't mind, I would like to work with radians instead of degrees.

First question a). The centre of the puck (let's call it $C$) is traveling at a constant speed, along a straight line which makes an angle of $\frac \pi4(=45^\circ)$ to the positive $x$-axis. This will be the line $y=x$. Now we start at the origin and we know that the point $\big(x(t), y(t)\big)$ has a distance of $24t$ inches from the origin. So we solve $$24 t= \sqrt{x(t)^2+y(t)^2}=\sqrt{2\cdot x(t)^2}=\sqrt 2 \,x(t).$$ This means $x(t)=y(t)=12\sqrt 2 \,t$. So the parametrization for the centre of the puck will be $$\begin{cases}x_C(t)=12\sqrt 2\, t\\ y_C(t)= 12\sqrt 2\, t\end{cases}$$

Now for question b). Now it is quite well known that for a point on the unit circle that spins around the origin, we have that $x=\cos \theta $ and $y=\sin \theta$, as according to this picture:

This can very easily be modified to a rotation around a point $(a,b)$ of a circle with radius $r$, like this: $x=a+r\cos \theta$ and $y=b+r\cos \theta$. Now apply this to the hockey puck. It rotates around the point $\big ( x_C(t),y_C(t)\big) $ and has radius $\frac 32$. We also know that it makes three rotations (that's $6\pi$ radians) per second and that our point (let's call it $P$) starts at $\frac\pi2(=90^\circ)$. So we get the parametrization: $$\begin{cases} x_P(t)=&x_C(t)+\frac32 \cos \left(6\pi t+\frac\pi 2\right) =&12\sqrt2\,t -\frac32\sin(6\pi t)\\ y_P(t)=&y_C(t)+\frac32 \sin\left(6\pi t+\frac \pi2\right )=& 12\sqrt 2\,t+ \frac32\cos (6\pi t)\end{cases} $$

Answer 2

(a) The hockey puck is traveling along the line $y=x$, traveling 24 inches every second. Since the puck moves 24 inches when $t=1$ and the puck travels at a constant pace, we can graph:

When $t=1$, $x = y = 12\sqrt{2}$. Therefore, the center of the puck at time $t$ is at: \begin{align*} x &= 12\sqrt{2}t \\ y &= 12\sqrt{2}t \end{align*}

(b) Let's make a parametric equation for the puck using variables: \begin{align*} x &= a + r \cos \theta \\ y &= b + r \sin \theta \end{align*} We know that the center of the puck at time $t$ is at $12 \sqrt{2} t$. We also know that the radius is $\frac{3}{2}$. Filling in the variables, we get: \begin{align*} x &= 12 \sqrt{2}t + \frac{3}{2} \cos \theta \\ y &= 12 \sqrt{2}t + \frac{3}{2} \sin \theta \end{align*} But what is $\theta$? Since the puck spins $6\pi$ radians every second and starts at $\frac{\pi}{2}$, $\theta = 6\pi t + \frac{\pi}{2}$ \begin{align*} x &= 12 \sqrt{2}t + \frac{3}{2} \cos (6\pi t + \frac{\pi}{2}) \\ y &= 12 \sqrt{2}t + \frac{3}{2} \sin (6\pi t + \frac{\pi}{2}) \end{align*} Since $\sin(\theta+90) = \cos(\theta)$ and $\cos(\theta + 90) = -\sin \theta$ \begin{align*} x &= 12 \sqrt{2}t - \frac{3}{2} \sin (6\pi t) \\ y &= 12 \sqrt{2}t + \frac{3}{2} \cos (6\pi t) \end{align*}