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Question Number 21967 by Tinkutara last updated on 07/Oct/17

$$\mathrm{A}\mathrm{block}\mathrm{is}\mathrm{tied}\mathrm{with}\mathrm{a}\mathrm{thread}\mathrm{of}\mathrm{length}l$$$$\mathrm{and}\mathrm{moved}\mathrm{in}\mathrm{a}\mathrm{horizontal}\mathrm{circle}\mathrm{on}\mathrm{a}$$$$\mathrm{rough}\mathrm{table}.\mathrm{Coefficient}\mathrm{of}\mathrm{friction}$$$$\mathrm{between}\mathrm{block}\mathrm{and}\mathrm{table}\mathrm{is}\mu =0.2.$$$$\mathrm{Find}\tan \theta ,\mathrm{where}\theta \mathrm{is}\mathrm{the}\mathrm{angle}$$$$\mathrm{between}\mathrm{acceleration}\mathrm{and}\mathrm{frictional}$$$$\mathrm{force}\mathrm{at}\mathrm{the}\mathrm{instant}\mathrm{when}\mathrm{speed}\mathrm{of}$$$$\mathrm{particle}\mathrm{is}v=\sqrt{1.6lg}$$

Commented by Tinkutara last updated on 07/Oct/17

Answered by ajfour last updated on 09/Oct/17

$$\tan \theta =\frac{v^2/l}{\mu g}=\frac{1.6g}{0.2g}=8.$$

Commented by Tinkutara last updated on 09/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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