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)) .


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Question Number 137787 by physicstutes last updated on 06/Apr/21

$A string A B of lenght 2 l has a particle attached to its midpoint C .$ $The ends A and B of the string are fastened to two fixed points$ $with A distance l vertically above B . With both parts of the string$ $taunt, the particle describes a horizontal circle about the line A B with$ $constant angular speed ω . If the tension in C A is three times$ $that in C B, prove that the angular velocity is 2 \sqrt{\frac{g}{l}} .$
	Answered by mr W last updated on 06/Apr/21

	Commented by mr W last updated on 06/Apr/21

	$A B = B C = C A = l$ $r = l \cos 30 °$ $T_{1} = 3 T_{2} a s g i v e n$ $i n Y - d i r e c t i o n :$ $T_{1} \sin 30 ° = m g + T_{2} \sin 30 °$ $3 T_{2} \sin 30 ° = m g + T_{2} \sin 30 °$ $2 T_{2} \sin 30 ° = m g$ $\Rightarrow T_{2} = m g$ $\Rightarrow T_{1} = 3 m g$ $i n X - d i r e c t i o n :$ $m r ω^{2} = T_{1} \cos 30 ° + T_{2} \cos 30 ° = 4 m g \cos 30 °$ $m l \cos 30 ° ω^{2} = 4 m g \cos 30 °$ $l ω^{2} = 4 g$ $\Rightarrow ω = 2 \sqrt{\frac{g}{l}}$
	Commented by physicstutes last updated on 06/Apr/21

	$this is fantastic sir, i had never thought this is$ $an equilaterial triangle . Thank you so much sir$
	Commented by mr W last updated on 07/Apr/21

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