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

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