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Question Number 66250 by Tanmay chaudhury last updated on 11/Aug/19

Answered by mr W last updated on 11/Aug/19

$$x_A=r\tan \theta$$$$u=\frac{{dx}_A}{dt}=\frac{r}{{\cos }^2\theta }\times \frac{d\theta }{dt}$$$$\Rightarrow \omega =\frac{d\theta }{dt}=\frac{u{\cos }^2\theta }{r}$$$$\Rightarrow \alpha =\frac{d\omega }{dt}=-\frac{u2\cos \theta \sin \theta }{r}\times \frac{d\theta }{dt}=-\frac{2u\sin \theta \cos \theta }{r}\times \frac{u{\cos }^2\theta }{r}$$$$\Rightarrow \alpha =-\frac{2u^2\sin \theta {\cos }^3\theta }{r^2}$$$$$$$$a_{Br}=r{\omega }^2=\frac{u^2{\cos }^4\theta }{r}$$$$a_{Bt}=r\alpha =-\frac{2u^2\sin \theta {\cos }^3\theta }{r}$$$$\Rightarrow a_B=\sqrt{a_{Br}^2+a_{Bt}^2}=\frac{u^2{\cos }^3\theta }{r}\sqrt{{\cos }^2\theta +{4\sin }^2\theta }$$$$\Rightarrow a_B=\frac{u^2{\cos }^3\theta }{r}\sqrt{1+{3\sin }^2\theta }$$$$\Rightarrow n=3$$

Commented by Tanmay chaudhury last updated on 11/Aug/19

$$excellentsir...$$

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