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Question Number 20819 by ajfour last updated on 03/Sep/17

Commented by ajfour last updated on 03/Sep/17

A particle moves along the spiral shown. Determine the magnitude of the velocity of the particle in terms of b, 𝛉, 𝛉^. .

$${A}\:{particle}\:{moves}\:{along}\:{the}\:{spiral} \\ $$$${shown}.\:{Determine}\:{the}\:{magnitude} \\ $$$${of}\:{the}\:{velocity}\:{of}\:{the}\:{particle}\:{in} \\ $$$${terms}\:{of}\:\boldsymbol{{b}},\:\boldsymbol{\theta},\:\overset{.} {\boldsymbol{\theta}}\:. \\ $$$$ \\ $$

Answered by dioph last updated on 04/Sep/17

(d/dt) rθ^2 = 0 r^. θ^2 + 2rθθ^. = 0 r^. = −((2rθθ^. )/θ^2 ) = − ((2rθ^. )/θ) r^. = −((2bθ^. )/θ^3 ) v = (√((r^. )^2 +r^2 (θ^. )^2 )) v = (√(((4b^2 )/θ^6 )(θ^. )^2 + (b^2 /θ^4 )(θ^. )^2 )) v = ((bθ^. )/θ^2 )(√((4/θ^2 )+1))

$$\frac{{d}}{{dt}}\:{r}\theta^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\overset{.} {{r}}\theta^{\mathrm{2}} \:+\:\mathrm{2}{r}\theta\overset{.} {\theta}\:=\:\mathrm{0} \\ $$$$\overset{.} {{r}}\:=\:−\frac{\mathrm{2}{r}\theta\overset{.} {\theta}}{\theta^{\mathrm{2}} }\:=\:−\:\frac{\mathrm{2}{r}\overset{.} {\theta}}{\theta} \\ $$$$\overset{.} {{r}}\:=\:−\frac{\mathrm{2}{b}\overset{.} {\theta}}{\theta^{\mathrm{3}} } \\ $$$${v}\:=\:\sqrt{\left(\overset{.} {{r}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \left(\overset{.} {\theta}\right)^{\mathrm{2}} } \\ $$$${v}\:=\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{2}} }{\theta^{\mathrm{6}} }\left(\overset{.} {\theta}\right)^{\mathrm{2}} \:+\:\frac{{b}^{\mathrm{2}} }{\theta^{\mathrm{4}} }\left(\overset{.} {\theta}\right)^{\mathrm{2}} } \\ $$$${v}\:=\:\frac{{b}\overset{.} {\theta}}{\theta^{\mathrm{2}} }\sqrt{\frac{\mathrm{4}}{\theta^{\mathrm{2}} }+\mathrm{1}} \\ $$

Commented by ajfour last updated on 04/Sep/17

yes sir, so it is.

$${yes}\:{sir},\:{so}\:{it}\:{is}. \\ $$

Commented by ajfour last updated on 04/Sep/17

what if the curve is r=be^(𝛉^2 /2) ?

$${what}\:{if}\:\:{the}\:{curve}\:{is}\:\:\boldsymbol{{r}}=\boldsymbol{{be}}^{\boldsymbol{\theta}^{\mathrm{2}} /\mathrm{2}} \:? \\ $$

Commented by dioph last updated on 04/Sep/17

r^. = be^(θ^2 /2) θθ^. v = (√((r^. )^2 +r^2 (θ^. )^2 )) v = (√(b^2 θ^2 e^θ^2 (θ^. )^2 +b^2 e^θ^2 (θ^. )^2 )) v = be^(θ^2 /2) θ^. (√(θ^2 +1))

$$\overset{.} {{r}}\:=\:{be}^{\theta^{\mathrm{2}} /\mathrm{2}} \:\theta\overset{.} {\theta} \\ $$$${v}\:=\:\sqrt{\left(\overset{.} {{r}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \left(\overset{.} {\theta}\right)^{\mathrm{2}} } \\ $$$${v}\:=\:\sqrt{{b}^{\mathrm{2}} \theta^{\mathrm{2}} {e}^{\theta^{\mathrm{2}} } \left(\overset{.} {\theta}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} {e}^{\theta^{\mathrm{2}} } \left(\overset{.} {\theta}\right)^{\mathrm{2}} } \\ $$$${v}\:=\:{be}^{\theta^{\mathrm{2}} /\mathrm{2}} \overset{.} {\theta}\sqrt{\theta^{\mathrm{2}} +\mathrm{1}} \\ $$

Commented by dioph last updated on 04/Sep/17

indeed, thanks for pointing out. it is now corrected

$$\mathrm{indeed},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{pointing}\:\mathrm{out}. \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{now}\:\mathrm{corrected} \\ $$

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