Question Number 80341 by Rio Michael last updated on 02/Feb/20
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{round}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{curve} \\ $$$${r}\:=\:{a}\left(\mathrm{1}\:+\:\mathrm{cos}\:\theta\right)\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular}\: \\ $$$$\mathrm{velocity}\:\omega\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{transverse}\:\mathrm{component} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}. \\ $$
Answered by mr W last updated on 02/Feb/20
Commented by mr W last updated on 02/Feb/20
$${r}={a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$\omega=\frac{{d}\theta}{{dt}}={constant} \\ $$$${v}_{{T}} ={v}_{\theta} =\frac{{rd}\theta}{{dt}}=\omega{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)={tranverse}\:{velocity} \\ $$
Commented by Rio Michael last updated on 02/Feb/20
$${thanks}\:{sir} \\ $$