A-particle-P-moving-at-constant-angular-velocity-describes-a-part-y-f-At-time-t-0-the-particle-is-at-the-point-with-coordinate-a-pi-2-and-moving-with-a-transverse-acceleration-of-2a-2

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Question Number 96761 by Rio Michael last updated on 04/Jun/20

$$\mathrm{A}\mathrm{particle}\mathrm{P}\mathrm{moving}\mathrm{at}\mathrm{constant}\mathrm{angular}\mathrm{velocity}$$$$\mathrm{describes}\mathrm{a}\mathrm{part}y=f\left(\theta \right).\mathrm{At}\mathrm{time}t=0,\mathrm{the}\mathrm{particle}$$$$\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{point}\mathrm{with}\mathrm{coordinate}(a,\frac{\pi }{2})\mathrm{and}\mathrm{moving}\mathrm{with}\mathrm{a}$$$$\mathrm{transverse}\mathrm{acceleration}\mathrm{of}-2a{\omega }^2\sin \theta .\mathrm{find}\mathrm{the}\mathrm{polar}\mathrm{equation}$$$$\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{described}\mathrm{by}\mathrm{this}\mathrm{particle}.\mathrm{Show}\mathrm{that}\mathrm{the}$$$$\mathrm{radial}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{acceleration}\mathrm{of}\mathrm{P}\mathrm{is}-a{\omega }^2(1+\cos \theta ).$$

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