Question Number 21967 by Tinkutara last updated on 07/Oct/17
$$\mathrm{A}\:\mathrm{block}\:\mathrm{is}\:\mathrm{tied}\:\mathrm{with}\:\mathrm{a}\:\mathrm{thread}\:\mathrm{of}\:\mathrm{length}\:{l} \\ $$$$\mathrm{and}\:\mathrm{moved}\:\mathrm{in}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{rough}\:\mathrm{table}.\:\mathrm{Coefficient}\:\mathrm{of}\:\mathrm{friction} \\ $$$$\mathrm{between}\:\mathrm{block}\:\mathrm{and}\:\mathrm{table}\:\mathrm{is}\:\mu\:=\:\mathrm{0}.\mathrm{2}. \\ $$$$\mathrm{Find}\:\mathrm{tan}\:\theta,\:\mathrm{where}\:\theta\:\mathrm{is}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{between}\:\mathrm{acceleration}\:\mathrm{and}\:\mathrm{frictional} \\ $$$$\mathrm{force}\:\mathrm{at}\:\mathrm{the}\:\mathrm{instant}\:\mathrm{when}\:\mathrm{speed}\:\mathrm{of} \\ $$$$\mathrm{particle}\:\mathrm{is}\:{v}\:=\:\sqrt{\mathrm{1}.\mathrm{6}{lg}} \\ $$
Commented by Tinkutara last updated on 07/Oct/17
Answered by ajfour last updated on 09/Oct/17
$$\mathrm{tan}\:\theta=\frac{{v}^{\mathrm{2}} /{l}}{\mu{g}}\:=\frac{\mathrm{1}.\mathrm{6}{g}}{\mathrm{0}.\mathrm{2}{g}}\:=\mathrm{8}\:. \\ $$
Commented by Tinkutara last updated on 09/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$