Three equal particles of mass m are fixed on a massless radial string at distances r, 2r and 3r from the centre. The system rotates with angular speed ω. If T₁, T₂ and T₃ are the tensions in the string segments from the centre outward, then T₁: T₂: T₃ is:
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Question

Three equal particles of mass m are fixed on a massless radial string at distances r, 2r and 3r from the centre. The system rotates with angular speed ω. If T₁, T₂ and T₃ are the tensions in the string segments from the centre outward, then T₁: T₂: T₃ is:

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G1 Physics · Accepted Answer

For the outermost mass at 3r, only T₃ acts inward, so T₃ = mω²(3r) = 3mω²r. For the middle mass at 2r, T₂ acts inward and T₃ outward, so T₂ − T₃ = mω²(2r). Thus T₂ = 2mω²r + 3mω²r = 5mω²r. For the inner mass at r, T₁ − T₂ = mω²r, giving T₁ = 6mω²r. Hence T₁:T₂:T₃ = 6:5:3.

Three equal particles of mass m are fixed on a massless radial string at distances r, 2r and 3r from the centre. The system rotates with angular speed ω. If T₁, T₂ and T₃ are the tensions in the string segments from the centre outward, then T₁:T₂:T₃ is: