Google interview riddle and scaling arguments

Question

I am puzzled by a riddle to which I have been told the answer and I have loads of difficulties to believe in the result.

The riddle goes as follows:

"imagine you are shrunk to the size of a coin (i.e. you are, say, scaled down by two orders of magnitude) but your density remains the same. You are put into a blender of hight 20 cm. The blender will start working in 60 seconds, what to you do?"

One of the best answers is apparently:

"I jump out of the blender to escape (yes the blender is still open luckily)."

This seems ultra non-intuitive to me and I have tried to find flaws in this answer but it seems to be fairly robust.

There are two ways you can think of it:

• the mass scales as $\sim \: L^3$ and therefore it will be $10^6$ times smaller. If we imagine that the takeoff velocity $v_{toff}$ is the same as before being rescaled $v_{big}$. We get then the hight at which a mini us can jump by equating the takeoff kinetic energy and the potential energy i.e. $mv_{toff}^2/2=mgh $ $\Rightarrow$ $h_{mini} = v^2_{big}/(2g) = h_{big} \sim 20 \rm cm$

• The second way to see it is to look more in details on how the power produced by muscles scales with the size of muscles. Basically, the power scales with the cross section of the muscle i.e. with the number of parallel "strings" pulling on the joints to contract the muscle. This implies that $P_{mini}=P_{big}/\alpha^2$ ($\alpha$ being the factor bigger than 1 by which you have been rescaled). We know that the takeoff kinetic energy will be given by $P \Delta t$. We assume now that $\Delta t \sim L/v_{big}$ so that $\Delta t_{mini} = L/(\alpha v_{big})$. In the end, this calculation tells us that $E_{mini} \sim E_{big}/\alpha^3$. However, equating again with the potential energy to get the hight we have $h_{mini} \sim E_{mini}/(m_{mini}g) = (E_{big}/\alpha^3)/(gm_{big}/\alpha^3)=E_{big}/(m_{big}g) = h_{big} \sim 20\:\rm cm$

These two reasonings seem fair enough and yet I don't trust the result they lead to. I would like to know if I am experiencing pure denial because of my prejudices or if there is some kind of flaw in the reasonings above (e.g. the fact that it is always assumed that the speed is unchanged when changing scale).

I also know that some tiny animals can jump more or less as high as human beings but it seems that most of the time these species have to use some kind of "trick" to store elastic energy in their body so as to generate enough kinetic energy at the takeoff to effectively jump super high.

If anyone has any thought on this, that would be very much welcome.

Answer 1

Let's assume our gravitational potential is zero at our center of mass just before the jump. Our initial mechanical energy is zero. We do nonconservative work to increase our mechanical energy. Then our feet leave the floor and our kinetic energy diminishes until we reach height $h$. We have $$W_{\mathrm{nc}} = F d = \frac{1}{2}m v^2 + m g d = m g h.$$ Therefore, $$h = \frac{F d}{m g}.$$

Strength is proportional to area and mass to volume so relative strength is inversely proportional to length, $F/m = k/L$. (Take $L$ to be our height, for example.) In addition, the distance we travel before leaving the ground is proportional to $L$, $d = k' L$. Therefore, $$h = \frac{k/L \times k'L}{g} = \mathrm{const}.$$ Thus, we will jump just as high as we had jumped before we were shrunk! Since a blender sized jump is a modest one for a normal man, we should be able to jump out without too much difficulty.

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Addendum: Here's a simple argument why power scales like force. For constant acceleration and zero initial velocity $v^2 = 2 a d$. Therefore, $F = m a = m v^2/(2 d)$ and so $$v = \sqrt{\frac{2 F d}{m}} \propto \sqrt{\frac{L^2\times L}{L^3}} = 1.$$ Therefore, instantaneous power scales like force, $$P = F v \propto L^2\times 1 = L^2.$$

A key fact regarding the scaling of force that may not be apparent is that we assume muscle fibers have a size that does not scale with $L$. (This is certainly true of humans of different heights and should make intuitive sense, smaller animals aren't made out of smaller molecules.) Thus, a leg with a smaller cross section has fewer fibers---the area of the cross section essentially counts the fibers.

Answer 2

Think about it like this: in order to get all lengths (i.e. your own length as well as the height of your jump) to scale down by a factor $\alpha$, while keeping the contraction velocity of your muscles the same, you have to rescale all lengths and all times occurring in the problem. That means that you have to scale down the gravity (length over time squared) by a factor $\alpha/\alpha^2= 1/\alpha$.

In other words: if you can jump up half your own heigth, after a scaling down in linear size by a factor 100, you would still be able to jump half your height on a planet with 100 times the gravitational acceleration of earth.

So effectively what is assumed is that you can arrive at a consistent physics when you scale down all length and all time scales by a factor $\alpha$, and your mass by a factor $\alpha^{-3}$. The dimensionless jump height (jump height divided by body height) remains invariant under this scaling down operation. However, the point is that in this scenario you also have to rescale the gravitational acceleration (length over time^2) by a factor $\alpha^{-1}$. It goes without saying that if you don't do this and keep the gravitational acceleration as weak as it is before scaling, you will be capable to jump much higher in terms of your own body height.

Answer 3

Many simplifying assumptions of our reality must be made to arrive at the results you're looking for (and that the other answers gave). I'll try to give my own 2 cents. I hope that anything incorrect in here will be corrected by other users - I'm mostly just trying to address some problems and confusion with these assumptions.

When you jump, you lower your center of mass by a distance $d$, and then your muscles contract and extend in various ways to push your center of mass up until you leave the ground. All the energy that gets expended in jumping gets converted to kinetic energy, $\frac{1}{2}mv^2$, and then that energy gets converted to potential energy, $mgh$. When you get to that height $h$, you are at the peak of your jump.

Your [striated] muscles are made up of sections called muscle fasicles, which contain many muscle fibers, and inside those muscle fibers there are many myofibrils (as well as other things, but the myofibrils are what we're interested in). Each myofibril can be thought of as a collection of pistons (sarcomeres) (with more details/complications that I will say in the next paragraph). In the end, your body is made up of a bunch of "pistons" (muscle fibers) that together, as an ensemble of pistons, make up one big piston, (your entire body).

When a "muscle" contracts and/or expands, what's happening is there are actually tiny systems known as myofibrils that make up your muscle fibers that are sliding and contracting their components to, in the big picture, relax or strain the muscle which they (the myofibrils make up).

This is where a problem arises. What exactly do we mean by scale down? Certainly you can't just scale down an atom without changing fundamental aspects of the universe you are considering. One can't just scale down myofibrils, as the nerves and proteins that they function by can't be scaled down without changing a fundamental aspect of the universe we are considering. Thus, the only notion of scaling down we can settle on is cutting down in amount. For a muscle, this literally means decreasing the number of muscle fibers.

From this assumption/consideration, we can see that the force that one can apply is essentially proportional to the cross-section of the muscle (i.e. the number of muscle fibers within the muscle), but not exactly. Since we've agreed that it is the number of muscle fibers that are decreasing, not their fundamental size/shape, a scale in the length of the myofibrils (and hence, and scale in length of the constituent sarcomeres) could definitely have an impact on their activity (i.e. maximum strength the pistons can apply). In fact, this is the case. Interesting non-linear behavior is seen between the length of the sarcomeres and their respective activity.

We've now run into another problem with our simplifying assumption that scaling down means reducing in number. However, if we step outside reality even further and assume that the activity of sarcomeres are not affected by their length, then the results from the other answers follow immediately. I'll state them now.

The work you do when you jump off the ground is given by $Fd$, where $F$ is the force you can apply and $d$ is the distance your center of mass covers with respect to your initial standing position. This will, assuming an ideal case, all be converted into kinetic energy which will all become potential energy when you reach the peak of your jump, at which your potential energy will be $mgh$, where $m$ is your mass, $g$ is the assumed constant acceleration downwards due to gravity, and $h$ is the maximum height you attain.

$$Fd=mgh \implies h=\frac{Fd}{mg}$$

If your volume is scaled down by a factor $k^3$ ($k$ in each direction of the canonical basis of $\mathbb{R}^3$, WLOG), then (assuming no change in mass-density) your mass will also scale down by $k^3$. By the assumptions made in the previous paragraphs, the force you apply to jump off the ground will scale down only by a factor $k^2$ (approximate scaling of the number of muscle fasicles), and obviously the distance that you displace your center of mass will scale by $k$. Thus,

$$h'=\frac{(k^2F)(kd)}{(k^3m)g}=\frac{Fd}{mg}=h$$

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I have omitted many calculations, such as how the non-linear change in the effective 'maximum piston force' affects the height you can jump. However, by reading that paper I referenced, I still think one would be able to "jump out of the blender", but I'm not completely certain - it may not be that simple. If anybody could add to this, it would be much appreciated.

Answer 4

The scaling arguments shown above are correct, which is consistent with the fact that animals of very different sizes - from a flea to an elephant - jump to the height on the order of 1 meter. Perhaps insects are different from mammals; so let's say a mouse vs. an elephant is a legitimate example of scaling (by about 100 in linear dimension) of a pretty similar organism - but both jump to about same height.

Answer 5

Well, this one's a bit less tedious and easier to understand. It all depends on how a blender actually works. The most common blenders have a couple of vertical knives as blades included. When these knives rotate, a tornado of air with a vortex is created . The vertical blades push the veggies upwards creating the vortex, the horizontal ones then slice off the veggies as soon as they get to the bottom of the vortex and the lowly bent blades make sure that none of the veggies are stuck at the bottom and left from grinding. I would try to climb and stick out on the top most part of inclination of the horizontal blades, which are slightly tilted, and will be acting like a slide for me. I'll then wait for the blender to start and as soon as it starts, I'll be thrown up with the vortex and finally out of the blender assuming that it's lid is open. P.S. I tried blending tomatoes with open lid, and believe me, the experience wasn't very tidy. I may surely break some bones while being thrown out of a vortex spinning at 300 RPM or more but it would certainly be better than getting juiced up.

Answer 6

Consider a muscle fibre as a thin rod of "spring constant" $K=YA/L$. $Y$ is modulus of elasticity. Energy stored as potential = $\frac{1}{2}$ stress $\times$ strain $\times$ volume. which is proportional to cube of length. Now jump height achieved = energy/ (mass * g) hence jump height remains unchanged!