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Question Number 74418 by peter frank last updated on 24/Nov/19

Answered by ajfour last updated on 24/Nov/19

$$\stackrel{{}^{\stackrel{N}{\nwarrow }}\underset{\underset{mg}{\downarrow }}{\mathbb{C}}{\to }_{{mv}^2/R}}{╱}$$$$╱$$$$\underset{\cdot ){-}^{\theta }-}{╱}$$$$mg\cos \theta +\frac{{mv}^2}{R}\sin \theta =N$$$$\mu N+mg\sin \theta =\frac{{mv}^2}{R}\cos \theta$$$$\Rightarrow \mu g\cos \theta +\frac{\mu v^2}{R}\sin \theta +g\sin \theta =\frac{v^2}{R}\cos \theta$$$$v^2=\frac{gR(\mu +\tan \theta )}{(1-\mu \tan \theta )}$$$$\Rightarrow v=\sqrt{\frac{gR(\mu +\tan \theta )}{(1-\mu \tan \theta )}}.$$

Commented by $@ty@m123 last updated on 24/Nov/19

$${}^{\nwarrow }\underset{\downarrow }{\mathbb{C}}\to$$$$╱$$$$╱$$$$╱$$$$\measuredangle \mathrm{\_}\theta \mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$Wow...$$$$Nicedrawingusingthisapp.$$$$Quitesurprising.$$$$Wherethereiswillthereisaway.$$$$$$

Commented by peter frank last updated on 24/Nov/19

$$thankyousirajfour$$

Commented by peter frank last updated on 07/Dec/19

$$pleasesirhelpQn74419$$

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