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A-string-AB-of-lenght-2l-has-a-particle-attached-to-its-midpoint-C-The-ends-A-and-B-of-the-string-are-fastened-to-two-fixed-points-with-A-distance-l-vertically-above-B-With-both-parts-of-the-string-

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Question Number 137787 by physicstutes last updated on 06/Apr/21
A string AB of lenght 2l has a particle attached to its midpoint C. The ends A and B of the string are fastened to two fixed points with A distance l vertically above B.With both parts of the string taunt,the particle describes a horizontal circle about the line AB with constant angular speed ω. If the tension in CA is three times that in CB, prove that the angular velocity is 2(√(g/l)) .
$$\mathrm{A}\:\mathrm{string}\:{AB}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{2}{l}\:\mathrm{has}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{its}\:\mathrm{midpoint}\:\mathrm{C}.\: \\ $$$$\mathrm{The}\:\mathrm{ends}\:{A}\:\mathrm{and}\:{B}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string}\:\mathrm{are}\:\mathrm{fastened}\:\mathrm{to}\:\mathrm{two}\:\mathrm{fixed}\:\mathrm{points} \\ $$$$\mathrm{with}\:{A}\:\mathrm{distance}\:{l}\:\mathrm{vertically}\:\mathrm{above}\:\mathrm{B}.\mathrm{With}\:\mathrm{both}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{the}\:\mathrm{string} \\ $$$$\mathrm{taunt},\mathrm{the}\:\mathrm{particle}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:{AB}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{angular}\:\mathrm{speed}\:\omega.\:\mathrm{If}\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:{CA}\:\mathrm{is}\:\mathrm{three}\:\mathrm{times} \\ $$$$\mathrm{that}\:\mathrm{in}\:{CB},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{angular}\:\mathrm{velocity}\:\mathrm{is}\:\mathrm{2}\sqrt{\frac{\mathrm{g}}{{l}}}\:. \\ $$
Answered by mr W last updated on 06/Apr/21
Commented by mr W last updated on 06/Apr/21
AB=BC=CA=l r=l cos 30° T_1 =3T_2 as given in Y−direction: T_1 sin 30°=mg+T_2 sin 30° 3T_2 sin 30°=mg+T_2 sin 30° 2T_2 sin 30°=mg ⇒T_2 =mg ⇒T_1 =3mg in X−direction: mrω^2 =T_1 cos 30°+T_2 cos 30°=4mg cos 30° ml cos 30°ω^2 =4mg cos 30° lω^2 =4g ⇒ω=2(√(g/l))
$${AB}={BC}={CA}={l} \\ $$$${r}={l}\:\mathrm{cos}\:\mathrm{30}° \\ $$$${T}_{\mathrm{1}} =\mathrm{3}{T}_{\mathrm{2}} \:{as}\:{given} \\ $$$${in}\:{Y}−{direction}: \\ $$$${T}_{\mathrm{1}} \:\mathrm{sin}\:\mathrm{30}°={mg}+{T}_{\mathrm{2}} \mathrm{sin}\:\mathrm{30}° \\ $$$$\mathrm{3}{T}_{\mathrm{2}} \:\mathrm{sin}\:\mathrm{30}°={mg}+{T}_{\mathrm{2}} \mathrm{sin}\:\mathrm{30}° \\ $$$$\mathrm{2}{T}_{\mathrm{2}} \:\mathrm{sin}\:\mathrm{30}°={mg} \\ $$$$\Rightarrow{T}_{\mathrm{2}} \:={mg} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\mathrm{3}{mg} \\ $$$${in}\:{X}−{direction}: \\ $$$${mr}\omega^{\mathrm{2}} ={T}_{\mathrm{1}} \mathrm{cos}\:\mathrm{30}°+{T}_{\mathrm{2}} \mathrm{cos}\:\mathrm{30}°=\mathrm{4}{mg}\:\mathrm{cos}\:\mathrm{30}° \\ $$$${ml}\:\mathrm{cos}\:\mathrm{30}°\omega^{\mathrm{2}} =\mathrm{4}{mg}\:\mathrm{cos}\:\mathrm{30}° \\ $$$${l}\omega^{\mathrm{2}} =\mathrm{4}{g} \\ $$$$\Rightarrow\omega=\mathrm{2}\sqrt{\frac{{g}}{{l}}} \\ $$
Commented by physicstutes last updated on 06/Apr/21
this is fantastic sir, i had never thought this is an equilaterial triangle. Thank you so much sir
$$\mathrm{this}\:\mathrm{is}\:\mathrm{fantastic}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{had}\:\mathrm{never}\:\mathrm{thought}\:\mathrm{this}\:\mathrm{is} \\ $$$$\mathrm{an}\:\mathrm{equilaterial}\:\mathrm{triangle}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 07/Apr/21

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