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# Question-66250

Question Number 66250 by Tanmay chaudhury last updated on 11/Aug/19
Answered by mr W last updated on 11/Aug/19
$${x}_{{A}} ={r}\:\mathrm{tan}\:\theta \\$$$${u}=\frac{{dx}_{{A}} }{{dt}}=\frac{{r}}{\mathrm{cos}^{\mathrm{2}} \:\theta}×\frac{{d}\theta}{{dt}} \\$$$$\Rightarrow\omega=\frac{{d}\theta}{{dt}}=\frac{{u}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{r}} \\$$$$\Rightarrow\alpha=\frac{{d}\omega}{{dt}}=−\frac{{u}\:\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{{r}}×\frac{{d}\theta}{{dt}}=−\frac{\mathrm{2}{u}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{{r}}×\frac{{u}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{r}} \\$$$$\Rightarrow\alpha=−\frac{\mathrm{2}{u}^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{3}} \:\theta}{{r}^{\mathrm{2}} } \\$$$$\\$$$${a}_{{Br}} ={r}\omega^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{4}} \:\theta}{{r}} \\$$$${a}_{{Bt}} ={r}\alpha=−\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{3}} \:\theta}{{r}} \\$$$$\Rightarrow{a}_{{B}} =\sqrt{{a}_{{Br}} ^{\mathrm{2}} +{a}_{{Bt}} ^{\mathrm{2}} }=\frac{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{3}} \:\theta}{{r}}\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{4sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{a}_{{B}} =\frac{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{3}} \:\theta}{{r}}\sqrt{\mathrm{1}+\mathrm{3sin}^{\mathrm{2}} \:\theta} \\$$$$\Rightarrow{n}=\mathrm{3} \\$$
Commented by Tanmay chaudhury last updated on 11/Aug/19
$${excellent}\:{sir}… \\$$