A cup made of $3/16$-inch-thick glass has an inside radius of $3$ inches and a total height of $6$ inches (including the bottom thickness of glass). The glass has a density of $165$ $lb/ft^3$. The jar is placed in water with a density of $62.5$ $lb/ft^3$. Assuming the jar remains upright, how far will the cup sink in the water?
Relatively mundane question for my intro to engineering class, probably a simple duck-up. This is the last problem on a homework assignment. I'm under remote learning because of COVID-19, so I can't talk to any of my classmates at the moment.
Here's how I went about solving it:
First, i calculated the volume of the glass shell:
$$ \begin {align} V_g &= \pi\left(r+\frac{3}{16}\right)^2(h) - \pi r^2\left(h-\frac{3}{16}\right) \\ &= \pi\left(3+\frac{3}{16}\right)^2(6) - \pi(3)^2\left(6-\frac{3}{16}\right) \\ &= 6\pi\left(\frac{51}{16}\right)^2 - 9\pi\left(\frac{93}{96}\right) \\ &= \frac{6\pi\times2601}{256} - \frac{9\pi\times93}{96} \\ &= \frac{15606\pi}{256} - \frac{837\pi}{96} \\ &= \frac{46818\pi}{768} - \frac{6696\pi}{768} \\ &= \frac{40122\pi}{768} (=164.1236 \text{ in}^3) \end {align} $$
Then I calculated the weight of the glass. Since the value was pounds per feet, I had to convert it to fit.
$$ \begin{align} W &= VD_w \\ &= \frac{164.1236}{1296}\times 165 \\ &= 0.0949789\times 165 \\ &= 15.67153 \text{ lb} \end{align} $$
Unless I'm missing something, the weight of water displaced should be equal to the weight of the cup, which is, again, $15.67153$.
Then, to calculate the volume of the water displaced, I run it backwards through the weight-volume conversion using the density of water this time.
$$ \begin{align} V_w & = \frac{W}{D_w} \times 1296\\ & = \frac{15.67153}{62.5} \times 1296\\ & = 324.9648715 \text{ in}^3 \end{align} $$
Since the cup is sitting upright, that would mean the volme displaced would be a cylinder, with a radius of the cup and height of the displacement! So, I plug in the volume and radius and solve for height.
$$ \begin{align} V &= \pi r^2 h \\ \frac{V}{\pi r^2} &= h \\ \frac{324.9648715}{\pi \left(\frac{51}{16}\right)^2} &= h \\ \frac{324.9648715}{31.91907223} &= h \\ 10.18 &= h...? \\ \end{align} $$
This... is... Impossible. The cup is $6$ inches tall, and will sink $10$ inches. There is no way a cup that of these dimensions will completely sink in the water! I have spent the last hour trying to solve this. More annoying is the fact that I have completely written out my process and the answer still makes no sense. Can someone point out my error?
You have a calculation mistake $(6-\frac{3}{16})=\frac{93}{16}$ and not $\frac{93}{96}$