- #1
clm222
So in advance I know that this has little practical application, but it was a fun math problem for a high school student. Earlier today I was bored and made myself a cup of tea. I poured some sugar in and thought to myself, "Given the height of the sugar glass (with a circular hole) from the surface of the tea, how much sugar per second is entering the tea?"
I thought about how I could calculate this, and decided that I could imagine the sugar particles in free fall, if the hole of the sugar glass is parallel to the surface of the tea I can simply find the volume of the cylinder in the drink and multiply by density to find mass. Of course the sugar won't fall in a cylinder after hitting the water, but the same amount of volume will dissolve as if there was a cylinder of it, right? That is, I treat the sugar as if there was no liquid, but find how much sugar is passing the point as if it was in free fall, I hope that makes sense.
If that's the case, the volume of the sugar will be \pi l{r^2}, where "l" is the length of the cylinder and "r" is the radius of the sugar glass. Now if the volume of the dissolved sugar is equal to the volume of the imaginary cylinder of sugar, than calculating the volume shouldn't be very difficult.
l={v_o}{t_d}+{\frac{1}{2}}g{t_d^2} (td= time of 'falling in drink', vo is velocity of sugar at the surface of the drink.)
{v_0}=g{t_{falling}} (h=height of sugar glass above drink) {v_0}=g\sqrt{\frac{2h}{g}}=\sqrt{2gh}
∴l={t_d}\sqrt{2gh}+{\frac{1}{2}}g{t_d^2}
now if we're looking at volume of sugar per second, td=1
∴l=\sqrt{2gh}+\frac{1}{2}g
the volume of the sugar 'cylinder'
V=\pi {r^2}(\sqrt{2gh}+\frac{1}{2}g)
multiply by density "ρ" to find "M", mass per second.
M=\pi \rho{r^2}(\sqrt{2gh}+\frac{1}{2}g)
Please let me know if my initial assumption was wrong, I'm only wondering the amount of sugar going in, not what happens to it. Besides, it seems like this would only approximate it. Please let me know if any of my math or physics was wrong, thanks a lot.
I thought about how I could calculate this, and decided that I could imagine the sugar particles in free fall, if the hole of the sugar glass is parallel to the surface of the tea I can simply find the volume of the cylinder in the drink and multiply by density to find mass. Of course the sugar won't fall in a cylinder after hitting the water, but the same amount of volume will dissolve as if there was a cylinder of it, right? That is, I treat the sugar as if there was no liquid, but find how much sugar is passing the point as if it was in free fall, I hope that makes sense.
If that's the case, the volume of the sugar will be \pi l{r^2}, where "l" is the length of the cylinder and "r" is the radius of the sugar glass. Now if the volume of the dissolved sugar is equal to the volume of the imaginary cylinder of sugar, than calculating the volume shouldn't be very difficult.
l={v_o}{t_d}+{\frac{1}{2}}g{t_d^2} (td= time of 'falling in drink', vo is velocity of sugar at the surface of the drink.)
{v_0}=g{t_{falling}} (h=height of sugar glass above drink) {v_0}=g\sqrt{\frac{2h}{g}}=\sqrt{2gh}
∴l={t_d}\sqrt{2gh}+{\frac{1}{2}}g{t_d^2}
now if we're looking at volume of sugar per second, td=1
∴l=\sqrt{2gh}+\frac{1}{2}g
the volume of the sugar 'cylinder'
V=\pi {r^2}(\sqrt{2gh}+\frac{1}{2}g)
multiply by density "ρ" to find "M", mass per second.
M=\pi \rho{r^2}(\sqrt{2gh}+\frac{1}{2}g)
Please let me know if my initial assumption was wrong, I'm only wondering the amount of sugar going in, not what happens to it. Besides, it seems like this would only approximate it. Please let me know if any of my math or physics was wrong, thanks a lot.