Question Number 6820 by Tawakalitu. last updated on 29/Jul/16 | ||
$${The}\:{position}\:{vector}\:{of}\:{the}\:{point}\:{P}\:\:{at}\:{time}\:{t}\:{is}\:{given}\:{by}\: \\ $$$$\left(\alpha{tant}\right){i}\:+\:\left(\alpha{sect}\right){k}\:{where}\:\alpha\:{is}\:{positive}\:{constant}\:{and} \\ $$$$\mathrm{0}\leqslant{t}\leqslant\frac{\Pi}{\mathrm{2}}.\:\:{Show}\:{that}\:{the}\:{velocity}\:{and}\:{acceleration}\:{of}\:{P}\:\:{when} \\ $$$${t}\:=\:\mathrm{0}\:{are}\:{at}\:{right}\:{angle}\:{to}\:{each}\:{other}.\:{If}\:{A}\:{is}\:{the}\:{point}\:{with}\: \\ $$$${position}\:{vector}\:\alpha{j},\:{obtain}\:{the}\:{vector}\:{equation}\:{for}\:{the}\:{straight}\: \\ $$$${line}\:{AP}\:{at}\:{time}\:{t}.\:\:{If}\:{the}\:{point}\:{Q}\:{divides}\:{AP}\:\:{internally}\:{in}\:{the} \\ $$$${ratio}\:\left({cost}\right)\::\:\left(\mathrm{1}\:−\:{cost}\right).\:{Show}\:{that}\:{the}\:{acceleration}\:{of}\:{the}\:{point} \\ $$$${Q}\:{is}\:{constant}\:{in}\:{magnitude}\:{and}\:{is}\:{always}\:{directed}\:{towards}\:{a}\: \\ $$$${fixed}\:{point}. \\ $$ | ||
Commented by Yozzii last updated on 29/Jul/16 | ||
$$\boldsymbol{{p}}\left({t}\right)=\begin{pmatrix}{\alpha{tant}}\\{\mathrm{0}}\\{\alpha{sect}}\end{pmatrix}\:\:\:\:\alpha={constant}>\mathrm{0},\:{t}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\Rightarrow\frac{{d}\boldsymbol{{p}}}{{dt}}\left({t}\right)=\boldsymbol{{v}}_{{p}} \left({t}\right)=\begin{pmatrix}{\alpha{sec}^{\mathrm{2}} {t}}\\{\mathrm{0}}\\{\alpha{sect}×{tant}}\end{pmatrix} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} \boldsymbol{{p}}}{{dt}^{\mathrm{2}} }\left({t}\right)=\boldsymbol{{a}}_{{p}} \left({t}\right)=\begin{pmatrix}{\mathrm{2}\alpha{sec}^{\mathrm{2}} {t}×{tant}}\\{\mathrm{0}}\\{\alpha\left({sect}×{tant}×{tant}+{sec}^{\mathrm{3}} {t}\right)}\end{pmatrix} \\ $$$$\boldsymbol{{a}}_{{p}} \left({t}\right)=\begin{pmatrix}{\mathrm{2}\alpha{tantsec}^{\mathrm{2}} {t}}\\{\mathrm{0}}\\{\alpha\left({tan}^{\mathrm{2}} {t}+{sec}^{\mathrm{2}} {t}\right){sect}}\end{pmatrix} \\ $$$$\boldsymbol{{v}}_{{p}} \left(\mathrm{0}\right)=\begin{pmatrix}{\alpha}\\{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:,\:\boldsymbol{{a}}_{{p}} \left(\mathrm{0}\right)=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\alpha}\end{pmatrix} \\ $$$$\therefore\:\boldsymbol{{v}}_{{p}} \left(\mathrm{0}\right)\bullet\boldsymbol{{a}}_{{p}} \left(\mathrm{0}\right)=\alpha×\mathrm{0}+\mathrm{0}×\mathrm{0}+\mathrm{0}×\alpha=\mathrm{0} \\ $$$${Since}\:\boldsymbol{{v}}_{{p}} \left(\mathrm{0}\right)\bullet\boldsymbol{{a}}_{{p}} \left(\mathrm{0}\right)=\mathrm{0},\:{the}\:{velocity} \\ $$$${and}\:{acceleration}\:{are}\:{perpendicular} \\ $$$${to}\:{each}\:{other}\:{at}\:{t}=\mathrm{0}. \\ $$$$\boldsymbol{{a}}\left({t}\right)=\begin{pmatrix}{\mathrm{0}}\\{\alpha}\\{\mathrm{0}}\end{pmatrix} \\ $$$${Direction}\:{vector}\:{of}\:{line}\:{AP}\:{is}\:{parallel}\:{to} \\ $$$$\boldsymbol{{b}}\left({t}\right)=\boldsymbol{{p}}\left({t}\right)−\boldsymbol{{a}}\left({t}\right)=\begin{pmatrix}{\alpha{tant}}\\{−\alpha}\\{\alpha{sect}}\end{pmatrix}=\alpha\begin{pmatrix}{{tant}}\\{−\mathrm{1}}\\{{sect}}\end{pmatrix} \\ $$$${So},\:{the}\:{vector}\:{equation}\:{of}\:{the}\:{line}\:{AP}\:{is} \\ $$$$\boldsymbol{{r}}=\begin{pmatrix}{\mathrm{0}}\\{\alpha}\\{\mathrm{0}}\end{pmatrix}\:+\mu\begin{pmatrix}{{tant}}\\{−\mathrm{1}}\\{{sect}}\end{pmatrix}\:\:\:\left(\mu\in\mathbb{R}\right). \\ $$$${For}\:{Q}\:{dividing}\:{AP}\:{in}\:{the}\:{ratio}\:\left({cost}\right):\left(\mathrm{1}−{cost}\right), \\ $$$$\mu={cost}\:\left({OQ}={OA}+{cost}\left({AP}\right)\right) \\ $$$$\therefore\:\boldsymbol{{q}}\left({t}\right)=\begin{pmatrix}{\mathrm{0}}\\{\alpha}\\{\mathrm{0}}\end{pmatrix}\:+{cost}\begin{pmatrix}{{tant}}\\{−\mathrm{1}}\\{{sect}}\end{pmatrix} \\ $$$$\boldsymbol{{q}}\left({t}\right)=\begin{pmatrix}{{sint}}\\{\alpha−{cost}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\boldsymbol{{v}}_{{q}} \left({t}\right)=\frac{{d}\boldsymbol{{q}}\left({t}\right)}{{dt}}=\begin{pmatrix}{{cost}}\\{{sint}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\boldsymbol{{a}}_{{q}} \left({t}\right)=\frac{{d}\boldsymbol{{v}}_{{q}} \left({t}\right)}{{dt}}=\begin{pmatrix}{−{sint}}\\{{cost}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\Rightarrow\mid\boldsymbol{{a}}_{{q}} \left({t}\right)\mid=\sqrt{\left(−{sint}\right)^{\mathrm{2}} +\left({cost}\right)^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} }=\mathrm{1} \\ $$$${Since}\:\mid\boldsymbol{{a}}_{{q}} \left({t}\right)\mid=\mathrm{1}\:{for}\:{all}\:{t}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right],\:{the} \\ $$$${acceleration}\:{of}\:{the}\:{point}\:{Q}\:{is}\:{constant}\:{in} \\ $$$${magnitude}. \\ $$$${Since}\:{the}\:{acceleration}\:{of}\:{Q}\:{has} \\ $$$${no}\:{component}\:{parallel}\:{to}\:{the}\:{unit}\:{vector}\:\boldsymbol{{k}}, \\ $$$${but}\:{variable}\:{components}\:{parallel}\:{to}\:\boldsymbol{{i}}\:{and}\:\boldsymbol{{j}}, \\ $$$$\boldsymbol{{a}}_{{q}} \left({t}\right)\:{acts}\:{entirely}\:{in}\:{the}\:{plane}\:{containing} \\ $$$${both}\:{the}\:{vectors}\:\boldsymbol{{i}}\:{and}\:\boldsymbol{{j}}.\: \\ $$$${Observe}\:{that}\:\boldsymbol{{v}}_{{q}} \left({t}\right)\bullet\boldsymbol{{a}}_{{q}} \left({t}\right)=\mathrm{0} \\ $$$$\Rightarrow{velocity}\:{and}\:{accleration}\:{of}\:{Q} \\ $$$${are}\:{normal}\:{to}\:{each}\:{other}. \\ $$$$ \\ $$ | ||
Commented by Tawakalitu. last updated on 30/Jul/16 | ||
$${Thanks}\:{so}\:{much}.\:{i}\:{appreciate}\:{your}\:{time} \\ $$ | ||