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In my physics book after this solved example:

A child of mass $m$ is initially at rest on top of a water slide at height h = 8.5m above the bottom of the slide. Assuming that the slide is frictionless because of water, find the speed of the child at bottom of slide.

a comment was written:

If we were asked to find the time taken for the child to reach the bottom of the slide, methods would be of no use; we would need to know the shape of the slide and we would have a difficult problem.

Why does the author say that we would need to know the shape of the slide to find the time taken for the child to reach bottom of the slide? Can't we use Newton's first law of motion in uniform acceleration to find the time?

we can find velocity at bottom $v$ = $\sqrt{2gh}$ = $13m/s$(approx) Using first law $v = u + at$ $13 = 0 + 9.8t$ $t = 13/9.8$

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    $\begingroup$ For the record, I want to single this question out as the right way to ask homework-like questions $\endgroup$ – Jim Jan 6 '15 at 14:49
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    $\begingroup$ You can certainly find the velocity at the bottom (though only if you assume almost all initial potential energy is converted to linear kinetic energy at the bottom, rather than being lost to heat due to friction during the trip). But that doesn't tell you the velocity as a function of time between top and bottom, since the shape of the slide creates a time-varying normal force in addition to gravity, so you can't assume "uniform acceleration". $\endgroup$ – Hypnosifl Jan 6 '15 at 19:37
  • $\begingroup$ There's a bit of differential math needed to find out the time for a non-trivial shape of the slide $x''(t) = slope(x(t)) \times g$ in other words the acceleration is dependent on the slope of where the child is sitting at the time. $\endgroup$ – ratchet freak Jan 6 '15 at 21:52
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    $\begingroup$ Simple thought experiment: If the slide initially has an infinitely shallow angle (ie, horizontal) then it will take an infinite amount of time for the "slidee" to reach the "knee" of the slide where descent actually begins. $\endgroup$ – Hot Licks Jan 7 '15 at 22:55
  • $\begingroup$ I think you answered your own question when you said "... law of motion in uniform acceleration ...". Whether or not you have uniform acceleration depends on the shape of the slide. $\endgroup$ – Dawood ibn Kareem Jan 8 '15 at 5:35
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Why does the author say that we would need to know the shape of the slide to find the time taken for the child to reach bottom of the slide?

As you've discovered, the speed going down a frictionless slide only depends on the vertical distance. This speed is not the vertical component of velocity. It is the magnitude of the velocity. The vertical component of velocity will be less than this on an inclined slide.

To make the geometry as simple as possible, I'll look at inclined ramps (no bumps, no curves; just a ramp at some angle inclined at with respect to horizontal). To keep the numbers simpler, I'll use g=10 m/s2 rather than 9.80665 m/s2. Suppose the slide has a vertical drop of 5 meters. That means the velocity at the bottom of the slide is 10 m/s. The average velocity is half that, 5 m/s.

Now let's put different length slides in place. A slide that is 5 meters long means you are falling rather than sliding. It takes one second to drop 5 meters. What if we used a ten meter long slide (i.e., inclined at a 30 degree angle with respect to horizontal). The velocity hasn't changed, but the distance has doubled. It takes two seconds to slide down this slide; twice as long as the vertical drop. Use an even longer slide, but still a 5 meter vertical drop, and it takes even longer to get to the bottom. With a 50 meter long slide (5.74 degrees with respect to horizontal), it takes ten seconds, or ten times as long to get to the bottom compared to the vertical drop.

In general, the time needed to reach the bottom of a frictionless inclined ramp is given by $t_\text{slide}=\frac l h t_\text{vert}$, where $l$ is the length of the ramp, $h$ is the vertical drop, and $t_\text{vert}$ is the time it takes to fall that same vertical distance.

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    $\begingroup$ I assumes the question is about slides of the same length, just different shape. $\endgroup$ – vsz Jan 8 '15 at 17:42
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Just for completeness, I'll explain how to obtain the time taken for an arbitrary curve.

If $h$ is the initial height of the child and $y$ the height once he has started falling. By energy conservation:

$$mgh=mgy+\frac{1}{2}mv^2\implies v=\sqrt{2g(h-y)}\tag{1} $$ We know know the speed at any time. Let us denote the horizontal position as $x$.

The distance traveled in a very small time interval can be written as:

$$ds=\sqrt{dx^2+dy^2}=\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy=\sqrt{1+x'^2}dy$$

So the speed is:

$$v=\frac{ds}{dt}=\sqrt{1+x'^2}\frac{dy}{dt}$$

Inserting this equation into $(1)$ and integrating leads to:

$$t=\frac{1}{\sqrt{2g}}\int_0^s \frac{\sqrt{1+x'^2}}{\sqrt{h-y}}dy $$

This integral gives you the time taken to reach the ground given any curve $y(x)$.

Furthermore, it is possible to obtain such curve, the tautochrone, that time taken is independent of the initial point:

enter image description here

Image source

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The shape of the slide definitely determines how long it takes to go down it. Consider if the slide was completely vertical. Now, a certain famous [recently deceased :( ] comedian had the astute observational powers to point out that this would, in fact, be a drop, not a slide. Nevertheless, you would quickly reach the bottom. Now imagine if the slide was like a rollercoaster; it went down, then back up, then down, etc. only reaching the ground at the very end. This up and down motion is purely a result of the shape of the slide and it must necessarily take longer than simply going straight down once. So you see, finding the time it takes to traverse the slide is heavily dependent on the shape of the slide

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  • $\begingroup$ @tom I was actually referring to John Pinette $\endgroup$ – Jim Jan 6 '15 at 14:39
  • $\begingroup$ but could a slide, that is not a drop, be frictionless? $\endgroup$ – Jodrell Jan 7 '15 at 14:03
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    $\begingroup$ @Jodrell I don't really see how that's relevant but Mag-lev slide? $\endgroup$ – Jim Jan 7 '15 at 14:05
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    $\begingroup$ A straight down slide is called a drop, a horizontal slide is called a shelf. Each has uses depending on how soon you want something to hit the floor. $\endgroup$ – Phil Jan 7 '15 at 23:45
  • $\begingroup$ @Jim "all physics problems take place in a vacuum anyway" indeed :P $\endgroup$ – Tobias Kienzler Jan 8 '15 at 10:14
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In your working you have assumed that $a=g$ - this is true if the slide is vertical.

Slides will have some angle, $\theta$ (e.g. $45^\circ$), which will mean that the acceleration, $a$ is given by $$a = g ~sin \theta$$

Note that $a$ will be less than $g$ because the value of the $sin$ term will be between $0$ and $1$. (except in the case of a vertical slide $\theta=90^\circ$ and $sin \theta=1$ so $a=g$).

But we are not done yet because most slides finish off horizontal and so there will be a curve in the slide at the bottom.

enter image description here

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    $\begingroup$ I think this'd be the quickest-to-digest response had you accompanied it with a diagram :) $\endgroup$ – Anti Earth Jan 7 '15 at 13:47
  • $\begingroup$ @AntiEarth - ok diagram coming up.... $\endgroup$ – tom Jan 7 '15 at 15:49
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In addition to the existing answers, it's worth noting that the fact that a slide slows down something sliding down it is actually how Galileo confirmed that objects fall at the same speed regardless of their mass. Just dropping them doesn't work well because things fall too fast to time, at least using Renaissance technology. So he built angled slides that would slow the falling down, allowing him to get better measurements of how long it was taking things to fall.

Also, imagine if things didn't work this way. We could build very efficient travel systems by building very slightly tilted surfaces: say, have it be ten feet high at one end, and end twenty miles away and five feet high. Then you'd be able to travel twenty miles in the same time it took you to fall five feet.

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The slide provides a normal force for the child. This alters the child's acceleration at various points on the slide thus affecting the time it takes to reach the bottom. The child is not free falling under constant uniform acceleration anymore, but instead follows the bumps. If the slide has a complicated enough shape then it would be difficult to find the time it will take to reach the bottom because of this variable net acceleration.

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  • $\begingroup$ I can understand that the acceleration is changing but how is the normal force altering the child's acceleration? $\endgroup$ – pcforgeek Jan 7 '15 at 1:07
  • $\begingroup$ Well if it's accelerating downwards with acceleration equal to $g$, and then suddenly the normal force acts to oppose gravity, the total acceleration will be $g - a$ where $a$ would be the acceleration due to that opposing force (the normal force). Think of an object on a table. It does not move because in that case the acceleration caused by the normal force is equal to $g$ hence the total acceleration of the system is $g - g = 0$ $\endgroup$ – Constandinos Damalas Jan 7 '15 at 1:10
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You know the final speed and you also know (because of conservation of kinetic + gravitational potential energy) that this is the maximum speed (at least, provided the slide stays above ground level). Call this maximum speed $v$.

For any time $t$, consider a (straight, shallow for large $t$) slide of length greater than $vt$. By the Mean Value Theorem or just by common sense, you cannot travel a distance greater than $vt$ in time $t$ without at some point exceeding speed $v$. So this particular slide takes some time longer than $t$ to travel.

So, the time taken to travel down the slide isn't just variable depending on the shape of the slide, it isn't even bounded above.

The error in your working is to take $a = 9.8$. This is true when falling, but it's not true when sliding down any shape other than a vertical cliff.

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protected by Qmechanic Jan 6 '15 at 19:15

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