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Question

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
mrω223
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B
mrω223
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C
mrω22+3
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D
mrω23
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Solution

The correct option is D mrω23
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).

T=mrω23T=mrω2T=mrω23

Hence option D is the correct answer.

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