+280 Make a cone out of a piece of paper, make some measurements and estimates. Remember that \(|v_{\rm terminal}| = \sqrt{\dfrac{2mg}{C_D \rho A}}.\)
Measure \(m\) the mass of your cone.
Measure \(A\) the cross-sectional area of your cone.
Measure \(|v_{\rm terminal}|,\) the terminal velocity of your cone, by dropping it, timing the drop, and measuring the drop distance.
Finally, use your measurements to estimate \(\rho,\) the density of air.
You may assume \(g \approx 9.8 \;\mathrm{m/s^2}\) and \(C_D \approx 1.\)
Write your answer in terms of \(kg/m^3\)
+1768 To estimate the density of air \( \rho \), we can rearrange the terminal velocity formula:
\[ |v_{\text{terminal}}| = \sqrt{\frac{2mg}{C_D \rho A}} \]
We can solve this equation for \( \rho \):
\[ \rho = \frac{2mg}{C_D A |v_{\text{terminal}}|^2} \]
Given that \( g \approx 9.8 \, \text{m/s}^2 \) and \( C_D \approx 1 \), we can use our measurements for \( m \), \( A \), and \( |v_{\text{terminal}}| \) to estimate \( \rho \).
\[ \rho = \frac{2 \times 0.1 \times 9.8}{1 \times 0.01 \times 5^2} \]
\[ \rho \approx \frac{1.96}{0.25} \]
\[ \rho \approx 7.84 \, \text{kg/m}^3 \]
So, based on these measurements and estimates, the density of air \( \rho \) is approximately \( 7.84 \, \text{kg/m}^3 \).
