Three particles are tied by means of strings of negligible mass and equal lengths. They move on a horizontal circular path of radius ′r′ as shown. Find the tension in the string connecting them.
A
mrω22−√3
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B
mrω22√3
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C
mrω22+√3
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D
mrω2√3
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Solution
The correct option is Dmrω2√3 Here by the free body diagram of single mass ,
All the strings will have same tension as all the strings are identical and connected to same mass at same distance.
Here by the FBD, analyzing a single mass will have two tensions at an angle of 60o as it forms an equilateral triangle, so by the law of vector addition the net resultant tension(T′) will be
T′=√T2+T2+2T2cos60∘ =√3T
The resultant tension will provide the necessary centripetal force (mω2r).