A uniform rod of length 2 m and density 7800 kg/m³ rotates in a horizontal plane about one end. The material has breaking stress 156 MPa. Neglect elastic extension while calculating the centrifugal stress. If the maximum safe angular speed is N rad/s, find N.
Physics · Properties of solid and fluid · properties of solid stress ,strain and modulus of elasticity
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Question

A uniform rod of length 2 m and density 7800 kg/m³ rotates in a horizontal plane about one end. The material has breaking stress 156 MPa. Neglect elastic extension while calculating the centrifugal stress. If the maximum safe angular speed is N rad/s, find N.

Physics · Properties of solid and fluid · properties of solid stress ,strain and modulus of elasticity

G1 Physics · Accepted Answer

For a uniform rod rotating about one end, the maximum tensile stress occurs at the rotating end. The maximum stress is σ(max) = ρω²L²/2. At the limiting safe speed, σ(max) = σ(breaking). Therefore, σ(breaking) = ρω(max)²L²/2. Hence, ω(max)² = [2σ(breaking)] / [ρL²]. Substitute: σ(breaking) = 156 MPa = 156×10⁶ Pa, ρ = 7800 kg/m³, L = 2 m. Thus, ω(max)² = [2×156×10⁶] / [7800×4]. The denominator is 31200. Therefore, ω(max)² = 10000. Hence, ω(max) = 100 rad/s. Thus, N = 100.