(Solved): mathematical methods for physicists solve in detail (a) Solve Example ...

mathematical methods for physicists

solve in detail

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(a) Solve Example 7.2.1, assuming that the parachute opens when the parachutist's velocity has reached \( v_{i}=60 \mathrm{mi} / \mathrm{h} \) (regard this time as \( t=0 \) ). Find \( v(t) \). (b) For a skydiver in free fall use the friction coefficient \( b=0.25 \mathrm{~kg} / \mathrm{m} \) and mass \( m=70 \mathrm{~kg} \). What is the limiting velocity in this case? We want to find the velocity of a falling parachutist as a function of time and are particularly interested in the constant limiting velocity, \( v_{0} \), that comes about by air drag, taken to be quadratic, \( -b v^{2} \), and opposing the force of the gravitational attraction, \( m g \), of the Earth on the parachutist. We choose a coordinate system in which the positive direction is downward so that the gravitational force is positive. For simplicity we assume that the parachute opens immediately, that is, at time \( t=0 \), where \( v(t)=0 \), our initial condition. Newton's law applied to the falling parachutist gives \[ m \dot{v}=m g-b v^{2}, \] where \( m \) includes the mass of the parachute. Chapter 7 Ordinary Differential Equations The terminal velocity, \( v_{0} \), can be found from the equation of motion as \( t \rightarrow \infty \); when there is no acceleration, \( \dot{v}=0 \), and \[ b v_{0}^{2}=m g, \quad \text { or } \quad v_{0}=\sqrt{\frac{m g}{b}} . \] It simplifies further work to rewrite Eq. (7.4) as \[ \frac{m}{b} \dot{v}=v_{0}^{2}-v^{2} . \] This equation is separable, and we write it in the form \[ \frac{d v}{v_{0}^{2}-v^{2}}=\frac{b}{m} d t . \] Using partial fractions to write \[ \frac{1}{v_{0}^{2}-v^{2}}=\frac{1}{2 v_{0}}\left(\frac{1}{v+v_{0}}-\frac{1}{v-v_{0}}\right), \] it is straightforward to integrate both sides of Eq. (7.5) (the left-hand side from \( v=0 \) to \( v \), the right-hand side from \( t=0 \) to \( t \) ), yielding \[ \frac{1}{2 v_{0}} \ln \frac{v_{0}+v}{v_{0}-v}=\frac{b}{m} t . \] Solving for the velocity, we have \[ v=\frac{e^{2 t / T}-1}{e^{2 t / T}+1} v_{0}=v_{0} \frac{\sinh (t / T)}{\cosh (t / T)}=v_{0} \tanh \frac{t}{T}, \] where \( T=\sqrt{m / g b} \) is the time constant governing the asymptotic approach of the velocity to its limiting value, \( v_{0} \). Inserting numerical values, \( g=9.8 \mathrm{~m} / \mathrm{s}^{2} \), and taking \( b=700 \mathrm{~kg} / \mathrm{m}, m=70 \mathrm{~kg} \), gives \( v_{0}=\sqrt{9.8 / 10} \approx 1 \mathrm{~m} / \mathrm{s} \approx 3.6 \mathrm{~km} / \mathrm{h} \approx 2.234 \mathrm{mi} / \mathrm{h} \), the walking speed of a pedestrian at landing, and \( T=\sqrt{m / b g}=1 / \sqrt{10 \cdot 9.8} \approx 0.1 \mathrm{~s} \). Thus, the constant speed \( v_{0} \) is reached within a second. Finally, because it is always important to check the solution, we verify that our solution satisfies the original differential equation: \[ \dot{v}=\frac{\cosh (t / T)}{\cosh (t / T)} \frac{v_{0}}{T}-\frac{\sinh ^{2}(t / T)}{\cosh ^{2}(t / T)} \frac{v_{0}}{T}=\frac{v_{0}}{T}-\frac{v^{2}}{T v_{0}}=g-\frac{b}{m} v^{2} . \] A more realistic case, where the parachutist is in free fall with an initial speed \( v(0)>0 \) before the parachute opens, is addressed in Exercise 7.2.16.