In class we derived ( v(m) ) for a rocket in free space ( left(vec{F}_{ ext {ext }}=0 ight) ) : [ v=v_{0}+v_{ ext {ex }} ln left(frac{m_{0}}{m} ight) ] This result did not depend on how ( m ) was "expended"; it is true for any ( m(t) ). We cannot directly integrate the above equation to get ( x(m) ) for any ( m(t) ), since ( v=dot{x} ) and the resulting differential equation has three variables: ( x, m ), and ( t ). Assume that ( m(t)=m_{0}-k t ), where ( k>0 ) is a constant, and that ( x_{0}=0 ). Find ( x(t) )
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x(t) = cos(t)+i sin(t) x(t) = cos(t)+i sin(t)+j*cos(2*t)+k*sin(2*t) Show more…
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We have $\frac{d m}{d t}=-\mu$ or, $d m=-\mu d t$ Integrating $$ \int_{m_{0}}^{m} d m=-\mu \int_{0}^{t} d t \text { or, } m=m_{0}-\mu t $$ As $\vec{u}=0$ so, from the equation of variable mass system : or, $$ \begin{gathered} \left(m_{0}-\mu t\right) \frac{d \vec{v}}{d t}=\vec{F} \text { or, } \frac{d \vec{v}}{d t}=\vec{w}=\vec{F} /\left(m_{0}-\mu t\right) \\ \int_{0}^{\vec{v}} d \vec{v}=\vec{F} \int_{0} \frac{d t}{\left(m_{0}-\mu t\right)} \end{gathered} $$ Hence $$ \vec{v}=\frac{\vec{F}}{\mu} \ln \left(\frac{m_{0}}{m_{0}-\mu t}\right) $$
Physical Fundamentals Of Mdchanics
Laws of Conservation of Energy, Momemtum, and Angular Momentum
According to the question, $\vec{F}=0$ and $\mu=-d m / d t$ so the equation for this system becomes, $$ m \frac{d \vec{v}}{d t}=\frac{d m}{d t} \vec{u} $$ As $d \vec{v} \uparrow \downarrow \vec{u} \quad$ so, $m d v=-u d m$. Integrating within the limits : $$ \frac{1}{u} \int_{0}^{v} d v=-\int_{m_{0}}^{m} \frac{d m}{m} \text { or } \frac{v}{u}=\ln \frac{m_{0}}{m} $$ Thus, $v=u \ln \frac{m_{0}}{m}$ As $d \vec{v} \uparrow \downarrow \vec{u}$, so in vector form $\vec{v}=-\vec{u} \ln \frac{m_{0}}{m}$
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