Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 16504 by Tinkutara last updated on 23/Jun/17

An object moves in a circular path with  a constant speed in the xy plane with  the centre at the origin. When the  object is at x = −2 m, its velocity is  −(4 m/s)j^∧ . Then objects velocity at  y = 2 m is  (1) 4 m/s i^∧   (2) (−4 m/s) i^∧   Using this data, find objects acceleration  when it is at y = 2 m  (1) 8 m/s^2  i^∧   (2) −8 m/s^2  j^∧

$$\mathrm{An}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{at}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{is}\:\mathrm{at}\:{x}\:=\:−\mathrm{2}\:\mathrm{m},\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$$−\left(\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\overset{\wedge} {{j}}.\:\mathrm{Then}\:\mathrm{objects}\:\mathrm{velocity}\:\mathrm{at} \\ $$$${y}\:=\:\mathrm{2}\:\mathrm{m}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{4}\:\mathrm{m}/\mathrm{s}\:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{4}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{i}} \\ $$$$\mathrm{Using}\:\mathrm{this}\:\mathrm{data},\:\mathrm{find}\:\mathrm{objects}\:\mathrm{acceleration} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:{y}\:=\:\mathrm{2}\:\mathrm{m} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{i}} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\overset{\wedge} {{j}} \\ $$

Answered by ajfour last updated on 23/Jun/17

Commented by ajfour last updated on 23/Jun/17

 +ve y intercept is (0, 2) instead    of (0,4) as in fig. above .

$$\:+{ve}\:{y}\:{intercept}\:{is}\:\left(\mathrm{0},\:\mathrm{2}\right)\:{instead}\: \\ $$$$\:{of}\:\left(\mathrm{0},\mathrm{4}\right)\:{as}\:{in}\:{fig}.\:{above}\:. \\ $$

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by sma3l2996 last updated on 23/Jun/17

v^∧ (θ)=2w(−sinθi^�^∧  +cosθj^∧ )  at x=−2m; θ=π  so : v^∧ (π)=2wj^∧ =−4j^∧   w=−2rad/s  v^∧ (θ)=4(sinθi^∧ −cosθj^∧ )  at y=2m ; θ=(π/2)  v^∧ ((π/2))=(4m/s)i^∧   ∗∗  a^∧ =(dv^∧ /dt)=4w(cosθi^∧ +sinθj^∧ )=−8(cosθi+sinθj)  at y=2m ; θ=(π/2)  so: a^∧ =(−8m/s^2 )j^∧

$$\overset{\wedge} {{v}}\left(\theta\right)=\mathrm{2}{w}\left(−{sin}\theta\hat {{i}}+{cos}\theta\overset{\wedge} {{j}}\right) \\ $$$${at}\:{x}=−\mathrm{2}{m};\:\theta=\pi \\ $$$${so}\::\:\overset{\wedge} {{v}}\left(\pi\right)=\mathrm{2}{w}\overset{\wedge} {{j}}=−\mathrm{4}\overset{\wedge} {{j}} \\ $$$${w}=−\mathrm{2}{rad}/{s} \\ $$$$\overset{\wedge} {{v}}\left(\theta\right)=\mathrm{4}\left({sin}\theta\overset{\wedge} {{i}}−{cos}\theta\overset{\wedge} {{j}}\right) \\ $$$${at}\:{y}=\mathrm{2}{m}\:;\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$$\overset{\wedge} {{v}}\left(\frac{\pi}{\mathrm{2}}\right)=\left(\mathrm{4}{m}/{s}\right)\overset{\wedge} {{i}} \\ $$$$\ast\ast \\ $$$$\overset{\wedge} {{a}}=\frac{{d}\overset{\wedge} {{v}}}{{dt}}=\mathrm{4}{w}\left({cos}\theta\overset{\wedge} {{i}}+{sin}\theta\overset{\wedge} {{j}}\right)=−\mathrm{8}\left({cos}\theta{i}+{sin}\theta{j}\right) \\ $$$${at}\:{y}=\mathrm{2}{m}\:;\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${so}:\:\overset{\wedge} {{a}}=\left(−\mathrm{8}{m}/{s}^{\mathrm{2}} \right)\overset{\wedge} {{j}} \\ $$

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com