Asked by shankhdeep mojumdar · 4 years ago
A particle of mass 1 kg is revolved in a horizontal circle by the string of length 30cm and cross section area is 1mm² then find max angular speed with which particle can revolve if breaking stress of string is 4.8x10â¶N/m²
m=1kg ,L=30cm=30/100m ,A=1mm^2=1×(10^-3)^2=1×10^-6m^2 stress of string means tension of string=4.8×10^6N/m^2 , wmax=? T=mv^2/r T=mw^2r^2/r (..v=wr) 4.8×10^6N/m^2=1kg×w^2×r 4.8×10^6N/m^2=1kg×w^2×√A/√3.14 (A=πr^2) w=0.10s-1
Sadaf Nayab · 4 years ago
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