A sphere made of rubber has a density of 1.00 g/cm3 and a radius of 8.00 cm. It falls through air of density 1.20 kg/m3 and has a drag coefficient of 0.500. What is its terminal speed (in m/s)

m = p_sphere*V A = pi*r^2 v_t = sqrt (2mg/Dp_airA)

The Attempt at a Solution

m = (870 kg/m^3) (4/3) pi (0.085 m) ^3 = 2.24 kg

A = pi (0.085 m) ^2 = 0.0227 m^2 v_t = sqrt[2 (2.24 kg) (9.80 m/s^2) / (0.500) (1.20 kg/m^3) (0.0227 m^3) ] = 56.8 m/s

Terminal speed = 1,826.51m/s

Explanation:

Volume of a sphere is given by: V=4/3pir^3

Where r is radius of sphere

V=4*3.142 * (8) ^2/3

V = 2144.66cm^3

Converting to meters

V=2144.66cm^3 * (1m^3 / 1*10^-6cm^3)

V = 2.145*10^-3m^3

Area of sphereA = pi (8) ^2

A = 3.142*64=210.6cm^3

Converting to meter

201cm^ * (1m/10000cm^2)

A=0.0210m^2

Given:

Density of shere = 1.00kg/m^3

Drag coefficient = 0.500

Mass of sphere=?

Density of sphere = mass of sphere / volume of shere

Mass = 2144.66cm^3*1.00kgcm^3

Mass = 2144.66kg

Terminal speed, VT = Sqrt (2mg) / (DpA)

VT = Sqrt (2 * (2144.66) * 9.8)) / (0.500*1.20*0.021)

VT = Sqrt (42035.34/0.0126)

VT=Sqrt (3,336,137.78)

VT = 1,826.51m/s