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Question Number 16401 by sushmitak last updated on 21/Jun/17

v=2i+2j+5k r=i+9jāˆ’8k Find š›š I can do ((rƗv)/r^2 )=š›š and i get š›š= ((61iāˆ’21jāˆ’16k)/(146)) but i dont get wƗr=v. why?

$$\boldsymbol{{v}}=\mathrm{2}\boldsymbol{{i}}+\mathrm{2}\boldsymbol{{j}}+\mathrm{5}\boldsymbol{{k}} \\ $$$$\boldsymbol{{r}}=\boldsymbol{{i}}+\mathrm{9}\boldsymbol{{j}}āˆ’\mathrm{8}\boldsymbol{{k}} \\ $$$$\mathrm{Find}\:\boldsymbol{\omega} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{do}\:\frac{\boldsymbol{{r}}Ɨ\boldsymbol{{v}}}{{r}^{\mathrm{2}} }=\boldsymbol{\omega} \\ $$$$\mathrm{and}\:\mathrm{i}\:\mathrm{get}\:\boldsymbol{\omega}=\:\frac{\mathrm{61}\boldsymbol{{i}}āˆ’\mathrm{21}\boldsymbol{{j}}āˆ’\mathrm{16}\boldsymbol{{k}}}{\mathrm{146}} \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{get}\:\boldsymbol{{w}}Ɨ\boldsymbol{{r}}=\boldsymbol{{v}}. \\ $$$${why}? \\ $$

Answered by ajfour last updated on 21/Jun/17

v^� =Ļ‰^� Ɨr^� (agreed) r^� Ɨv^� = r^� Ɨ(Ļ‰^� Ɨr^� ) = r^2 Ļ‰^� āˆ’r^� (Ļ‰^� .r^� ) If Ļ‰^� .r^� ā‰  0 , r^� Ɨv^� ā‰  r^2 Ļ‰^� .

$$\:\bar {{v}}=\bar {\omega}Ɨ\bar {{r}}\:\:\:\:\left({agreed}\right) \\ $$$$\:\bar {{r}}Ɨ\bar {{v}}\:=\:\bar {{r}}Ɨ\left(\bar {\omega}Ɨ\bar {{r}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{r}^{\mathrm{2}} \:\bar {\omega}āˆ’\bar {{r}}\left(\bar {\omega}\:.\bar {{r}}\right)\: \\ $$$$\:\:{If}\:\:\:\bar {\omega}.\bar {{r}}\:\neq\:\mathrm{0}\:,\:\:\:\:\:\bar {{r}}Ɨ\bar {{v}}\:\neq\:{r}^{\mathrm{2}} \:\bar {\omega}\:. \\ $$

Commented by prakash jain last updated on 21/Jun/17

v=wƗr This gives the component of v which is perpendicular to r. In the given question vāˆ™r=(2+18āˆ’40)=āˆ’20ā‰ 0 component of v parallel to r =((vāˆ™r)/r^2 )r=āˆ’((20)/(1^1 +9^2 +(āˆ’8)^2 ))(i+9jāˆ’8k) =āˆ’((20)/(146))(i+9jāˆ’8k) comoponent of v perpendicular to r=vāˆ’((vāˆ™r)/r^2 )r =(2i+2i+5k)+(((20)/(146))i+((180)/(146))jāˆ’((160)/(146))k) =((312i+472j+570k)/(140)) āˆ’āˆ’āˆ’āˆ’āˆ’ v=š›šĆ—r=(1/(146))(61iāˆ’21jāˆ’16k)Ɨ(i+9jāˆ’8k) =(1/(146))(312i+472j+570k) which is the expected result. Also note that only tangential component of velocity contributes to angular velocity.

$$\boldsymbol{{v}}=\boldsymbol{{w}}Ɨ\boldsymbol{{r}} \\ $$$$\mathrm{This}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{component}\:\mathrm{of}\:\boldsymbol{\mathrm{v}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\boldsymbol{\mathrm{r}}. \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{given}\:\mathrm{question} \\ $$$$\boldsymbol{\mathrm{v}}\centerdot\boldsymbol{\mathrm{r}}=\left(\mathrm{2}+\mathrm{18}āˆ’\mathrm{40}\right)=āˆ’\mathrm{20}\neq\mathrm{0} \\ $$$$\mathrm{component}\:\mathrm{of}\:\boldsymbol{\mathrm{v}}\:\mathrm{parallel}\:\mathrm{to}\:\boldsymbol{\mathrm{r}} \\ $$$$=\frac{\boldsymbol{\mathrm{v}}\centerdot\boldsymbol{\mathrm{r}}}{\mathrm{r}^{\mathrm{2}} }\boldsymbol{\mathrm{r}}=āˆ’\frac{\mathrm{20}}{\mathrm{1}^{\mathrm{1}} +\mathrm{9}^{\mathrm{2}} +\left(āˆ’\mathrm{8}\right)^{\mathrm{2}} }\left(\boldsymbol{\mathrm{i}}+\mathrm{9}\boldsymbol{\mathrm{j}}āˆ’\mathrm{8}\boldsymbol{\mathrm{k}}\right) \\ $$$$=āˆ’\frac{\mathrm{20}}{\mathrm{146}}\left(\boldsymbol{\mathrm{i}}+\mathrm{9}\boldsymbol{\mathrm{j}}āˆ’\mathrm{8}\boldsymbol{\mathrm{k}}\right) \\ $$$$\mathrm{comoponent}\:\mathrm{of}\:\boldsymbol{\mathrm{v}}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\boldsymbol{\mathrm{r}}=\boldsymbol{\mathrm{v}}āˆ’\frac{\boldsymbol{\mathrm{v}}\centerdot\boldsymbol{\mathrm{r}}}{\mathrm{r}^{\mathrm{2}} }\boldsymbol{\mathrm{r}} \\ $$$$=\left(\mathrm{2}\boldsymbol{\mathrm{i}}+\mathrm{2}\boldsymbol{\mathrm{i}}+\mathrm{5}\boldsymbol{\mathrm{k}}\right)+\left(\frac{\mathrm{20}}{\mathrm{146}}\boldsymbol{\mathrm{i}}+\frac{\mathrm{180}}{\mathrm{146}}\boldsymbol{\mathrm{j}}āˆ’\frac{\mathrm{160}}{\mathrm{146}}\boldsymbol{\mathrm{k}}\right) \\ $$$$=\frac{\mathrm{312}\boldsymbol{\mathrm{i}}+\mathrm{472}\boldsymbol{\mathrm{j}}+\mathrm{570}\boldsymbol{\mathrm{k}}}{\mathrm{140}} \\ $$$$āˆ’āˆ’āˆ’āˆ’āˆ’ \\ $$$$\boldsymbol{\mathrm{v}}=\boldsymbol{\omega}Ɨ\boldsymbol{\mathrm{r}}=\frac{\mathrm{1}}{\mathrm{146}}\left(\mathrm{61}\boldsymbol{\mathrm{i}}āˆ’\mathrm{21}\boldsymbol{\mathrm{j}}āˆ’\mathrm{16}\boldsymbol{\mathrm{k}}\right)Ɨ\left(\boldsymbol{\mathrm{i}}+\mathrm{9}\boldsymbol{\mathrm{j}}āˆ’\mathrm{8}\boldsymbol{\mathrm{k}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{146}}\left(\mathrm{312}\boldsymbol{\mathrm{i}}+\mathrm{472}\boldsymbol{\mathrm{j}}+\mathrm{570}\boldsymbol{\mathrm{k}}\right) \\ $$$$\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{expected}}\:\boldsymbol{\mathrm{result}}. \\ $$$$\mathrm{Also}\:\mathrm{note}\:\mathrm{that}\:\mathrm{only}\:\mathrm{tangential} \\ $$$$\mathrm{component}\:\mathrm{of}\:\mathrm{velocity}\:\mathrm{contributes} \\ $$$$\mathrm{to}\:\mathrm{angular}\:\mathrm{velocity}. \\ $$

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