Particles each of mass ′m′ are tied by means of strings of equal length as shown in figure. If the system is rotated with constant angular velocity ′ω′ in horizontal plane, find the tension in the string.
[Given ′r′is the radius of the circle]
A
mrω2√2
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B
mrω2√5
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C
mrω22
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D
mrω22√2
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Solution
The correct option is Amrω2√2 Free body diagram of single mass (D) ,
From FBD, a single mass will have two tensions at an angle of 90o as it forms a square, so by the law of vector addition the net resultant tension(T′) will be
T′=√T2+T2+2T2cos900=T√2 and the resultant tension T′will provide the necessary centripetal force.
Hence T′=mrω2 ⇒T√2=mrω2⇒T=mrω2√2