Question Number 66250 by Tanmay chaudhury last updated on 11/Aug/19
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Answered by mr W last updated on 11/Aug/19
![x_A =r tan θ u=(dx_A /dt)=(r/(cos^2 θ))×(dθ/dt) ⇒ω=(dθ/dt)=((u cos^2 θ)/r) ⇒α=(dω/dt)=−((u 2 cos θ sin θ)/r)×(dθ/dt)=−((2u sin θ cos θ)/r)×((u cos^2 θ)/r) ⇒α=−((2u^2 sin θ cos^3 θ)/r^2 ) a_(Br) =rω^2 =((u^2 cos^4 θ)/r) a_(Bt) =rα=−((2u^2 sin θ cos^3 θ)/r) ⇒a_B =(√(a_(Br) ^2 +a_(Bt) ^2 ))=((u^2 cos^3 θ)/r)(√(cos^2 θ+4sin^2 θ)) ⇒a_B =((u^2 cos^3 θ)/r)(√(1+3sin^2 θ)) ⇒n=3](https://www.tinkutara.com/question/Q66252.png)
$${x}_{{A}} ={r}\:\mathrm{tan}\:\theta \\ $$$${u}=\frac{{dx}_{{A}} }{{dt}}=\frac{{r}}{\mathrm{cos}^{\mathrm{2}} \:\theta}×\frac{{d}\theta}{{dt}} \\ $$$$\Rightarrow\omega=\frac{{d}\theta}{{dt}}=\frac{{u}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{r}} \\ $$$$\Rightarrow\alpha=\frac{{d}\omega}{{dt}}=−\frac{{u}\:\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{{r}}×\frac{{d}\theta}{{dt}}=−\frac{\mathrm{2}{u}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{{r}}×\frac{{u}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{{r}} \\ $$$$\Rightarrow\alpha=−\frac{\mathrm{2}{u}^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{3}} \:\theta}{{r}^{\mathrm{2}} } \\ $$$$ \\ $$$${a}_{{Br}} ={r}\omega^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{4}} \:\theta}{{r}} \\ $$$${a}_{{Bt}} ={r}\alpha=−\frac{\mathrm{2}{u}^{\mathrm{2}} \mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{3}} \:\theta}{{r}} \\ $$$$\Rightarrow{a}_{{B}} =\sqrt{{a}_{{Br}} ^{\mathrm{2}} +{a}_{{Bt}} ^{\mathrm{2}} }=\frac{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{3}} \:\theta}{{r}}\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{4sin}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow{a}_{{B}} =\frac{{u}^{\mathrm{2}} \mathrm{cos}^{\mathrm{3}} \:\theta}{{r}}\sqrt{\mathrm{1}+\mathrm{3sin}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow{n}=\mathrm{3} \\ $$
Commented by Tanmay chaudhury last updated on 11/Aug/19
![excellent sir...](https://www.tinkutara.com/question/Q66255.png)
$${excellent}\:{sir}… \\ $$