Water slide

Even though this question sounds funny, it is actually very serious. From first hand experience at water parks in earlier years I know that there are kids out there who will do anything to quench the need for speed. For instance,

• taking a running start to the drop-in slide
• entering the ride with $n$ people on board the family tube and then having $m<n$ of the original passengers jump off backwards to boost velocity in the $\hat{z}$ direction, where $\hat{z}$ points roughly down the center of the chute
• etc.

...the point is, folks are scrappy when it comes to risking life and limb on water rides. It is therefore very important to protect extreme kids from the extreme danger of launching out the side of the rides and I am thankful for Dr. Mattingly for his guidance in calculating the prolonged existence of children the world over.

The thrill-seeking child has two effects competing for his attention. One is the force of gravity, which pulls him Earthward and tends to keep him within the confines of the chute, and the other is the centripetal force which seeks to launch him on a most impressive yet rather unfortunate trajectory.

Gravity accelerates him with magnitude $g$ straight down. Call the angle between the horizontal and the wall of the ride $\theta$ . The component of gravitational acceleration along the slant of the wall is given by $\displaystyle g\sin\theta$ .

The centripetal acceleration points outward along the radius of the turn with magnitude $\displaystyle \frac{mv^2}{R+x}$ where $R$ is the radius of the turn (from the center of the circle to the center line of the chute) and $x$ is the horizontal distance from the center line of the chute to the thrill-seeking child. The component of the centripetal acceleration $a_c$ along the slant of the wall is given by $\displaystyle a_c\cos\theta$ .

For the child to remain in the ride, these two forces have to balance, i.e.

$\displaystyle g\sin\theta = \frac{mv^2}{R+x}\cos\theta$

We know that the wall has the shape of the parabola $y = \frac12 x^2$ , so $\frac{dy}{dx} = x$ . If we draw a triangle to represent this information, the vertical $dy$ is given by $xdx$ , the horizontal $dx$ is given by $dx$ (no surprises here) and the hypotenuse is given by $\displaystyle\sqrt{1+x^2}dx$ . Therefore, we have

$\displaystyle \sin\theta = \frac{x}{\sqrt{1+x^2}}$

$\displaystyle \cos\theta = \frac{1}{\sqrt{1+x^2}}$

Plugging this into the balance condition we found above, we have

$\displaystyle g\frac{x}{\sqrt{1+x^2}} = \frac{v^2}{R+x}\frac{1}{\sqrt{1+x^2}}$

Solving for $x$ we find $\displaystyle\boxed{v = \sqrt{gRx+gx^2}}$

Finally, the maximum height of the walls is 1 m which occurs at a distance $x = \sqrt2$ m from the center line of the chute. Plugging this in to the relation, we find that the top speed of the thrill-seeking child before sailing over the wall for the plunge into the sharp reaches of the bottomless chasm is $\approx 9.43$ m/s .

We are only interested in the behavior at maximum height (1m). The radius at this height is $R = 5+\sqrt{2}$ . Applying standard motion laws, we have $\frac{mv^2}{R}\cos\theta=mg\sin\theta$ . Here, $\tan\theta=\frac{\partial}{\partial x}\left(x^2 \over 2\right)$ . At $x=\sqrt2$ , we have $\tan\theta=\sqrt2$ . Finally we have, $v^2=(5+\sqrt2)*9.8*\sqrt2=9.428^2$ .

If the child is travelling at speed $v$ around the curve at a height $y=\tfrac12u^2$ above the base of the slide, then the child is travelling along a circular path of radius $5+u$ . Let $N$ be the normal reaction between the child and the slide, and let $\theta$ be the angle that the tangent of the cross-section of the slide makes with the horizontal. Let the child's mass be $m$ .

Resolving forces vertically, $mg = N\cos\theta$ . Resolving forces horizontally/radially, $N\sin\theta = mv^2(5+u)^{-1}$ . Hence $v^2 \; = \; g(5+u)\tan\theta$ But $\tan\theta \,=\, \tfrac{dy}{dx}\Big|_{x=u} \,=\, u$ , and hence $v^2 \; = \; g(5+u)u$ Since the maximum height of the slide is $1$ m, we deduce that the maximum speed is $v_{\mathrm{max}} \; = \; \sqrt{g(5+\sqrt{2})\sqrt{2}} \; = \; 9.428 \mbox{ m s}^{-1}$

Let,

r = Radius of the horizontal circle
V = Linear velocity of the point mass

If y = 0.5 x^2 is the geometry of cross section of the water slide

Then radius of the circular path of the point mass in motion is R = (r + x) = (r + sqrt(2*y))

To maintain a constant vertical position at a height (y )during the motion sum of vertical forces must be zero.

List of forces acting:

Centrifugal force acting radially, Fc = (m * V^2)/R
Gravitational Force acting vertically downward = Fg = m * g (g = acceleration due to gravity)

If Fr is the resultant force of Fc and Fg acting at an angle theta to the vertical force Fg on the parabolic surface, then

Fr = (m * V^2) / (R * sin (theta))

then an equal force is applied by the parabolic surface on the particle in the opposite direction which is (-Fr) (Newton's third law)

Let this opposite force have a Vertical upper component (-Fu)

Then angle between (-Fu) and (-Fr) is theta (opposite angle are equal)

So, Fu = Fr * cos (theta)

To maintain this position during motion algebraic sum of vertical forces must vanish

Fu - Fg = 0

===> Fr * cos (theta) = m * g
===> (m * V^2) / (R * sin (theta)) * cos (theta) = m * g
===>V^2/ (R * tan (theta)) = g

or V = sqrt ( R * g * tan (theta) )
tan (theta) at (x,y) = dy/dx = x

Given, y = 1m, therefore x = sqrt(2*y) = sqrt(2) metre

R = (r + x) = (5 + sqrt(2)) metre
g = 9.8 m/s^2

V = 9.4284921 m/s

It is obvious that, the kid will attain the highest velocity at the farthest tip of the parabolic curve, from the center of curvature of the circular section. So, when the height is $1m.$ , the horizontal distance is $\sqrt{2}m.$ , and the slope at the farthest tip when $(x,y)=(\sqrt{2}m.,1m.)$ (considering the $x-axis$ of the parabola directed away from the center of curvature of the circular section) is $(t)$ can be calculated as follows: $y=\frac{x^2}{2}$ $\Rightarrow\frac{dy}{dx}=x\Rightarrow\frac{dy}{dx}|_{(\sqrt{2},1)}=t=\sqrt{2}.$ Also, the total radius $(r)$ , when the kid is at the farthest tip is $(5+\sqrt{2})m.$ The maximum velocity that can be attained can be expressed as $\sqrt{r*g*t}.$ Putting the respective values, we get, $\sqrt{\{5+\sqrt{2}\}*9.8*\sqrt{2}}=\fbox{9.42849m.}$ where $g=9.8m/s^2$ (acting downwards).

The faster the kid travels, the farther it is from the centre. Therefore we can canculate the highest speed by the equlibrium condition when the kid locates on the edge of the slide. At the edge of the slide, the sum of gravity and centrifugal force must be normal to the tangent at that point. Therefore we have an equation: \frac{dy}{dx} = \frac {v^2/(R+x)}{g} 2x = \frac {v^2/(R+x)}{g} x is calculated by x= \sqrt{2y} Ultimately, we get v= 9.428