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

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

Answered by ajfour last updated on 23/Jun/17

Commented by ajfour last updated on 23/Jun/17

$$+veyinterceptis(0,2)instead$$$$of(0,4)asinfig.above.$$

Commented by Tinkutara last updated on 24/Jun/17

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

Answered by sma3l2996 last updated on 23/Jun/17

$$\stackrel{\wedge }{v}\left(\theta \right)=2w(-sin\theta \hat{i}+cos\theta \stackrel{\wedge }{j})$$$$atx=-2m;\theta =\pi$$$$so:\stackrel{\wedge }{v}\left(\pi \right)=2w\stackrel{\wedge }{j}=-4\stackrel{\wedge }{j}$$$$w=-2rad/s$$$$\stackrel{\wedge }{v}\left(\theta \right)=4(sin\theta \stackrel{\wedge }{i}-cos\theta \stackrel{\wedge }{j})$$$$aty=2m;\theta =\frac{\pi }{2}$$$$\stackrel{\wedge }{v}\left(\frac{\pi }{2}\right)=\left(4m/s\right)\stackrel{\wedge }{i}$$$$\ast \ast$$$$\stackrel{\wedge }{a}=\frac{d\stackrel{\wedge }{v}}{dt}=4w(cos\theta \stackrel{\wedge }{i}+sin\theta \stackrel{\wedge }{j})=-8(cos\theta i+sin\theta j)$$$$aty=2m;\theta =\frac{\pi }{2}$$$$so:\stackrel{\wedge }{a}=(-8m/s^2)\stackrel{\wedge }{j}$$

Commented by Tinkutara last updated on 24/Jun/17

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

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