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Thread: Newtonian Mechanics

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    Newtonian Mechanics

    A skydiver jumps off a plane. The only force acting on the skydiver is her weight, inthis free fall. However, when the parachute opens, there is a resistive force ofmagnitude 𝛼𝑣2 newtons, where 𝑣 ms-1 is the speed and 𝛼 is a positive constantdepending on the size and construction of the parachute. The skydiver and parachuteare to be modelled throughout as a particle. The total mass of the skydiver and theparachute is 70 kg. Assume the acceleration due to gravity is 10 ms-2.
    Consider the parachute is open now. Apply the resistive force to findexpressions for the total distance travelled by the skydiver in terms of t and 𝛼.

    I am only able to figure out ma = mg - 𝛼𝑣2

    I've tried integrating 3 different equations to get the total displacement but on second thoughts i think it is wrong as t will not hold true for all 3 equations. I will greatly appreciate some help so i can continue from there and try to figure out the answer. Thanks in advance..
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    Re: Newtonian Mechanics

    $70v'(t) = 700-\alpha (v(t))^2$ is a second-order ordinary nonlinear differential equation.

    Simplify it:

    $\dfrac{v'}{ 10 \left( 1- \left( v \sqrt{ \dfrac{ \alpha }{ 700 } } \right)^2 \right) } = 1$

    Integrate both sides with respect to $t$. For the substitution, let $v\sqrt{\dfrac{\alpha}{700}} = \sin \theta$.

    You will need $\sec \theta = \sqrt{\dfrac{700}{700-\alpha v^2}}$ and $\tan \theta = \sqrt{\dfrac{\alpha v^2}{700-\alpha v^2}}$
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    Re: Newtonian Mechanics

    sry but how do i integrate the left side of the simplification? do i like get v/10(cosθ)2​ ? or should i take it as sin t when integrating?
    Last edited by noobpronoobpro; May 13th 2017 at 12:02 PM.
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    Re: Newtonian Mechanics

    $\displaystyle \int \dfrac{v'(t)dt}{10\left( 1-\left(v(t)\sqrt{\dfrac{\alpha}{700} } \right)^2 \right) } = \int dt$

    $v(t)\sqrt{\dfrac{\alpha }{700}} = \sin \theta$

    $v'(t) \sqrt{\dfrac{ \alpha }{700} } dt = \cos \theta d\theta$

    Plugging in the substitution:

    $\displaystyle \begin{align*}\int \dfrac{\cos \theta \sqrt{ \dfrac{700}{\alpha}} d\theta}{10\left(1-\sin^2 \theta \right)} & = t+C_1 \\ \sqrt{\dfrac{7}{\alpha} }\int \sec \theta d\theta & = t+ C_1 \\ \sqrt{\dfrac{7}{\alpha} } \ln \left| \sec \theta + \tan \theta \right| & = t+C_1\end{align*}$

    Now, reverse the substitution, solve for $v(t)$ and integrate again to get $s(t)$.
    Last edited by SlipEternal; May 13th 2017 at 12:15 PM.
    Thanks from noobpronoobpro
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    Re: Newtonian Mechanics

    sry to ask, but this equation only holds true for the total distance travelled, if he opened the parachute right at the start? In other words, the equation is calculating the total distance travelled after opening the parachute?
    Last edited by noobpronoobpro; May 13th 2017 at 07:13 PM.
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    Re: Newtonian Mechanics

    Yes
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    Re: Newtonian Mechanics

    okay thanks for the help ^^
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    Re: Newtonian Mechanics

    Quote Originally Posted by noobpronoobpro View Post
    okay thanks for the help ^^
    I guess I should not have stopped where I did. This problem is actually a little trickier than that.

    From where I left off, you had:
    $\sqrt{\dfrac{700}{700-\alpha v^2}} + \sqrt{\dfrac{\alpha v^2}{700-\alpha v^2}} = e^{\sqrt{\tfrac{\alpha}{7}}t + C_1}$

    Multiplying both sides by the conjugate gives:

    $1 = e^{\sqrt{\tfrac{\alpha}{7}}t + C_1}\left(\sqrt{\dfrac{700}{700-\alpha v^2}} - \sqrt{\dfrac{\alpha v^2}{700-\alpha v^2}} \right)$

    $\sqrt{\dfrac{700}{700-\alpha v^2}} - \sqrt{\dfrac{\alpha v^2}{700-\alpha v^2}} = e^{-\left(\sqrt{\tfrac{\alpha}{7}}t +C_1\right)}$

    Adding this together with the first formula gives:

    $2\sqrt{\dfrac{700}{700-\alpha v^2}} = e^{\sqrt{\tfrac{\alpha}{7}}t + C_1}+e^{-\sqrt{\tfrac{\alpha}{7}}t - C_1}$

    $\dfrac{700}{700-\alpha v^2} = \cosh^2\left(\sqrt{\dfrac{\alpha}{7}} t + C_1\right)$

    $1-\dfrac{\alpha}{700}v^2 = \text{sech}^2\left(\sqrt{\dfrac{\alpha}{7}}t + C_1\right)$

    $v(t) = 10\sqrt{\dfrac{7}{\alpha}}\tanh\left(\sqrt{\dfrac{ \alpha}{7}}t + C_1\right)$

    Integrating this, you get:

    $s(t) = \dfrac{70}{\alpha}\ln\left(\cosh\left(\sqrt{\dfrac {\alpha}{7}}t+C_1\right)\right)+C_2$

    Now, this just gives you the position function. For total distance, you will need:

    $\displaystyle \dfrac{70}{\alpha}\int_0^t { \left[ \ln \left( \cosh \left( \sqrt{ \dfrac{ \alpha }{ 7 } } z + C_1 \right) \right) + C_2 \right] dz }$

    I am not sure if there is a better solution. I am not a physicist, so for all I know, there is a much easier formula.
    Last edited by SlipEternal; May 15th 2017 at 11:59 AM.
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    Re: Newtonian Mechanics

    $\dfrac{dv}{dt} = g-\dfrac{\alpha}{m}v^2$

    note that terminal speed occurs when $\dfrac{dv}{dt} = 0 \implies |v_T| = \sqrt{\dfrac{mg}{\alpha}}$

    $\dfrac{dv}{dt} = g\left(1-\dfrac{\alpha}{mg}v^2\right)$

    let $\beta = \dfrac{\alpha}{mg}$ ...

    $\dfrac{dv}{dt} = g(1-\beta v^2)$

    $\dfrac{dv}{(1+\sqrt{\beta} v)(1-\sqrt{\beta} v)} = g \, dt$

    $\dfrac{1}{2}\bigg[\dfrac{dv}{1+\sqrt{\beta} v} + \dfrac{dv}{1-\sqrt{\beta} v} \bigg] = g \, dt$

    $\dfrac{\sqrt{\beta}}{1+\sqrt{\beta} v} \, dv - \dfrac{-\sqrt{\beta}}{1-\sqrt{\beta} v} \, dv = 2\sqrt{\beta} g \, dt$

    $\dfrac{1+\sqrt{\beta} v}{1-\sqrt{\beta} v} = Ae^{2\sqrt{\beta} g t}$

    ... if the chute is deployed prior to any appreciable acceleration $\implies v(0) = 0 \implies A = 1$

    $v = \dfrac{1}{\sqrt{\beta}} \bigg[\dfrac{1 - e^{-2\sqrt{\beta} gt}}{1 + e^{-2\sqrt{\beta} gt}} \bigg]$

    note ...

    $\displaystyle \lim_{t \to \infty} \dfrac{1}{\sqrt{\beta}} \bigg[\dfrac{1 - e^{-2\sqrt{\beta} gt}}{1 + e^{-2\sqrt{\beta} gt}} \bigg] = \dfrac{1}{\sqrt{\beta}} = \sqrt{\dfrac{mg}{\alpha}} = |v_T|$

    ... just fyi, if the chute is deployed after free-falling for a time $t_1$, then $v(t_1) = gt_1$, and the initial condition will change the value of the constant, $A$. I'm not going any further with this scenario.


    distance equation ...

    let $k = 2\sqrt{\beta} g$ ...

    $\dfrac{dx}{dt} = \dfrac{1}{\sqrt{\beta}} \bigg[\dfrac{1 - e^{-kt}}{1 + e^{-kt}} \bigg] = \dfrac{1}{\sqrt{\beta}} \bigg[\dfrac{1 + e^{-kt} - 2e^{-kt}}{1 + e^{-kt}} \bigg]$

    $\displaystyle x = \sqrt{\dfrac{mg}{\alpha}} \int_0^t 1 + \dfrac{-2e^{-kz}}{1 + e^{-kz}} \, dz$

    $x = \sqrt{\dfrac{mg}{\alpha}} \bigg[z + \dfrac{2}{k} \ln(1+e^{-kz})\bigg]_0^t = \sqrt{\dfrac{mg}{\alpha}} \bigg[t + \dfrac{2}{k} \ln \left(\dfrac{1+e^{-kt}}{2}\right) \bigg]$

    $k = 2\sqrt{\beta} g = 2 \sqrt{\dfrac{\alpha g}{m}}$

    $x = \sqrt{\dfrac{mg}{\alpha}} \cdot t + \dfrac{m}{\alpha} \cdot \ln \left(\dfrac{1+e^{- 2 \sqrt{\frac{\alpha g}{m}}t}}{2}\right)$

    this should be ok if I didn't make an algebra error somewhere ... it happens.
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