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This is a question I came up with while watching my friend squirt ketchup onto his table. He was squirting the ketchup out of a bottle while moving the bottle upwards.

A ketchup bottle starts upside down with the tip at the table. Ketchup is squirted out at a volume flow rate of $Q(t)$, in $\frac{m^3}{s}$, while the bottle itself is moving upwards at a rate of $v(t$), in $\frac{m}{s}$.

When the ketchup comes out of the bottle it is always initially not moving, but immediately starts falling to the table due to gravity ($g = 10 \frac{m}{s^2}$ downwards).

Find $V(t)$, the volume of ketchup on the table as a function of time.

Edit 1: We can ignore the fact that in real life the accumulating ketchup on the table will, in some sense, increase the height of the table.

enter image description here

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    $\begingroup$ Thumbs up because of the bizarreness of your question :) $\endgroup$ – Gonzalo Benavides Jul 4 '18 at 2:27
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    $\begingroup$ +1 for the graphics! $\endgroup$ – JavaMan Jul 4 '18 at 2:30
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    $\begingroup$ I'd like to thank your friend for inspiring this absurd question. $\endgroup$ – Carser Jul 4 '18 at 2:30
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    $\begingroup$ Are you having to squeeze the bottle or is it flowing out naturally? $\endgroup$ – Mohammad Zuhair Khan Jul 4 '18 at 2:31
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    $\begingroup$ @MohammadZuhairKhan, right, having a known geometry would be fine. Although, I'm not sure that any given geometry is much more accurate than assuming no height... $\endgroup$ – Carser Jul 4 '18 at 2:39
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If we let $x=0$ be the table top and measure upwards, the bottle position is $x(t)=vt$.
Ketchup that leaves the bottle at time $t$ takes $t'$ to fall where $vt=\frac 12g(t')^2$
Ketchup that leaves the bottle at time $t$ hits the table at $t+\sqrt{\frac {2vt}g}$
At time $u$ the ketchup that just arrived at the table left the bottle at time $t$ where $u=t+\sqrt{\frac {2vt}g}$
We would like to invert this equation $$u=t+\sqrt{\frac {2vt}g}\\(u-t)^2=\frac {2vt}g\\t^2-2ut-\frac {2v}gt+u^2=0\\ t=\frac 12\left(2u+\frac {2v}g-\sqrt{(2u+\frac {2v}g)^2-4u^2}\right)$$ and the amount of ketchup at time $u$ is $Q$ times this. If $Q$ is not a constant, you need to integrate $$\int_0^{t(u)}Q(t)dt$$to get the amount on the table at time $u$

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    $\begingroup$ This seems to be for a constant velocity $v(t)$, I think. What about when $v(t)$ is not constant? $\endgroup$ – Carser Jul 4 '18 at 3:32
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    $\begingroup$ @Carser: yes, I assumed constant $v$. If $v$ is not constant the same logic works. You need to look back in time to the time when the bottle was as a height that the ketchup that left is just hitting the table. If $v(t)$ is not simple you will have to find a numeric solution. $\endgroup$ – Ross Millikan Jul 4 '18 at 3:40

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