Let\'s look at the falling mouse example we touched on in class in more detail.
ID: 1592228 • Letter: L
Question
Let's look at the falling mouse example we touched on in class in more detail. Let's make things as simple as possible: consider a spherical mouse of radius 0.02 m and a spherical human of radius 0.25 m. These unfortunate creatures have the same density as water, 1000 kg/m^3. The volume of a sphere is V_sphere = 4/3 pi r^3. What are the masses of the human and the mouse? What are the weights of the human and the mouse? In this particular case, the approximate force of air resistance is given by the following formula, which I will never ask you about again: F_drag = 1/8 Times Density of air Times (Speed)^2 Times Surface Area of Sphere The density of air is 1.225 kg/m^3, and the surface area of a sphere is given by the formula SA_sphere = 4 pi r^2. At terminal speed, the forces on each animal are balanced that is, W = F_drag. Using all of this, find the terminal speed of both the mouse and the human.Explanation / Answer
a) Mass = volume*density
Mass = (4/3)*pi*r^3 * density
Mh = (4/3)*pi*(0.25)^3*1000 = 65.45 kg
Mm = (4/3)*pi*(0.2)^3*1000 = 0.0335 kg
b)
The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg
So, Wh = 65.45*9.8 = 641.41 kg
Wm = 0.0335 * 9.8 = 0.328 kg
c) W = Fdrag
or Wh = 1/8 * 1.225 * (4*pi*(0.25^2)) * (Tv)^2
So, 641.41 = 1/8 * 1.225 * (4*pi*(0.25^2)) * (Tv)^2
or Tv(human) = 73.03 m/s
And
Wm = 1/8 * 1.225 * (4*pi*(0.02^2)) * (Tv)^2
Tv(mouse) = 20.64 m/s
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