Question

An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container has a mass of 3.00 kg. The container has a drag coefficient of 0.500 and a diameter of 32.7 cm. With the unknown object inside, the container reaches a terminal velocity of 65.7 m/s. What is the mass of the unknown object? Use 1.22 kg/m ​3 ​​ for the density of air.

Answer 1

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

$F_D =F_g$

By definition we know that the Drag force is defined as

$F_D= \frac{1}{2} C_d \rho A V^2$

Where,

$C_d =$ Drag coefficient

$\rho =$Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

$F_g = (m_1 +m_2) g$

Where,

$m_1 =$Mass of sphere

$m_2 =$Mass of unknown object

Equating the two equations we have to

$(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2$

Re-arrange for m_2,

$m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1$

Our values are given by,

$C_d = 0.5$

$\rho = 1.22Kg/m^3$

$V = 66.7m/s$

$m_1 = 3Kg$

$d= 32.7*10^{-2}m$

$r = 16.35*10^{-2}m$

Replacing in the equation we have,

$m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3$

$m_2 = 8.62Kg$

Therefore the mass of unknown object is 8.62Kg