Question Number 91654 by Rio Michael last updated on 02/May/20
![th position vector of a particle p of mass 3 kg is given by r = [(cos 2t) i + (sin 2t)j] m given that p was intitialy at rest. find the cartesian equation of its path and describe it.](Q91654.png)
$$\mathrm{th}\:\mathrm{position}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:{p}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{3}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\: \\ $$$$\:{r}\:=\:\left[\left(\mathrm{cos}\:\mathrm{2}{t}\right)\:{i}\:+\:\left(\mathrm{sin}\:\mathrm{2}{t}\right){j}\right]\:\mathrm{m} \\ $$$$\mathrm{given}\:\mathrm{that}\:{p}\:\:\mathrm{was}\:\mathrm{intitialy}\:\mathrm{at}\:\mathrm{rest}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}\:\mathrm{and}\:\mathrm{describe}\:\mathrm{it}. \\ $$
Answered by Kunal12588 last updated on 02/May/20
![m = 3kg r = [(cos 2t)i+(sin 2t)j] m y=cos 2t x=sin 2t x^2 +y^2 =1 i.e. path is a circle](Q91658.png)
$${m}\:=\:\mathrm{3}{kg} \\ $$$${r}\:=\:\left[\left({cos}\:\mathrm{2}{t}\right){i}+\left({sin}\:\mathrm{2}{t}\right){j}\right]\:{m} \\ $$$${y}={cos}\:\mathrm{2}{t} \\ $$$${x}={sin}\:\mathrm{2}{t} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\:\:{i}.{e}.\:{path}\:{is}\:{a}\:{circle} \\ $$
Commented by Rio Michael last updated on 02/May/20
![wow thanks sir, what if we have something like this r = [ (cos 2t) i + (sin 2t)j + (sin 2t)k] how do describe this?](Q91659.png)
$$\mathrm{wow}\:\mathrm{thanks}\:\mathrm{sir}, \\ $$$$\mathrm{what}\:\mathrm{if}\:\mathrm{we}\:\mathrm{have}\:\mathrm{something}\:\mathrm{like}\:\mathrm{this} \\ $$$$\:{r}\:=\:\left[\:\left(\mathrm{cos}\:\mathrm{2}{t}\right)\:{i}\:+\:\left(\mathrm{sin}\:\mathrm{2}{t}\right){j}\:+\:\left(\mathrm{sin}\:\mathrm{2}{t}\right){k}\right]\:\mathrm{how}\:\mathrm{do}\:\mathrm{describe}\:\mathrm{this}? \\ $$
Commented by Kunal12588 last updated on 02/May/20
$${may}\:{be}\:{answer}\:{lies}\:{in}\:\mathrm{3}{d}\:{geometry},\:{I}\:{don}'{t}\:{know} \\ $$
Commented by Rio Michael last updated on 02/May/20
$$\mathrm{okay}\:\mathrm{sir}\:\mathrm{thanks},\:\mathrm{was}\:\mathrm{thinking}\:\mathrm{of}\:\mathrm{hw}\:\mathrm{to}\:\mathrm{describe}\:\mathrm{this} \\ $$