Respuesta :
1) [tex]161 \frac{lbf \cdot s}{ft}[/tex]
The particle experiences two forces during its fall:
- The force of gravity, [tex]mg[/tex], pointing downward, where m is the mass of the particle and g is the acceleration of gravity
- The drag force, [tex]kv[/tex], pointing upward, where k is a constant and v is the instantaneous speed of the particle
When the particle reaches its terminal speed, the acceleration becomes zero: this means that the net force is also zero, so the two forces are balanced. Therefore we can write
[tex]mg = kv_t[/tex]
where [tex]v_t[/tex] is the terminal velocity. We have the following data:
[tex]m = 1\cdot 10^{-13} slug= 1\cdot 10^{-13} \frac{lbf \cdot s^2}{ft}[/tex] is the mass
[tex]g = 32.2 ft/s^2[/tex] is the acceleration
[tex]v_t = 0.2 ft/s[/tex] is the terminal speed
Solving for k, we find
[tex]k=\frac{mg}{v_t}=\frac{(1 \frac{lbf \cdot s^2}{ft})(32.2 \frac{ft}{s^2})}{0.2 \frac{ft}{s}}=161 \frac{lbf \cdot s}{ft}[/tex]
2) 28.6 ms
We now want to find the time at which the particle reaches 99 % of the terminal speed, so a speed of
[tex]v' = 0.99 v_t = (0.99)(0.2)=0.198 ft/s[/tex]
The net force acting on the particle is
[tex]F=mg-kv[/tex]
So the acceleration is
[tex]a=\frac{dv}{dt}=g-\frac{k}{m}v[/tex]
This differential equation can be rewritten as
[tex]\frac{dv}{g-\frac{k}{m}v}=dt[/tex]
And by integrating on both sides and performing the calculation, we can get the final expression for the velocity as
[tex]v=\frac{mg}{k}(1-e^{-\frac{k}{m}t})[/tex]
We want to find the time t at which v = 0.198 ft/s, so we now need to re-arrange the formula for t. This gives:
[tex]t = -\frac{m}{k} ln (\frac{g-\frac{k}{m}v}{g})[/tex]
And substituting all the values, we find:
[tex]t = -\frac{1}{161} ln (\frac{32.2-\frac{161}{1}(0.198)}{32.2})=0.0286 s = 28.6 ms[/tex]
The value of the constant K and the time required to reach 99% of the terminal speed are; 161 lbf.t²/ft and 0.0286 s
What is the time required?
We are given;
Mass; m_g = 1 × 10⁻¹³ slugs = 1 × 10⁻¹³ lbf.t²/ft
terminal speed; V = 0.2 ft/s
acceleration due to gravity; g = 32.2 ft²/s
A) Thus, from FD = Kv, we know that mg = Fd. Thus;
mg = kV
k = mg/V
k = (1 × 10⁻¹³ × 32.2)/0.2
k = 161 lbf.t²/ft
B) At 99% of terminal speed, we now have;
New speed; v' = 99% * 0.2 = 0.198 ft/s
The net force is;
F= mg - kV
Since F = ma, then;
ma = mg - kV
Thus;
a = g - (k/m)V
By use of differential equation, we can get;
t = -(m/k)In(g - (k/m)v)/g)
t = -(1 × 10⁻¹³/161)In(32.2 - (161/(1 × 10⁻¹³))0.198)/32.2)
t = 0.0286 s
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