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A disc of mass m and radius r is gently placed on another disc of mass 2m and radius r. The disc of mass 2m is rotating with angular velocity `omega_(0)` initially. The disc is placed such that axis of both are concident. The coefficient of friction is `mu` for surfaces of contact Assume that pressure on disc is uniformely distributed. Loss in kinetic energy of systemA. `(1)/(6)mr^(2)omega_(0)^(2)`B. `(1)/(3)mr^(2)omega_(0)^(2)`C. `(1)/(2)mr^(2)omega_(0)^(2)`D. None of these |
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Answer» Correct Answer - A By COE `Mg.(L)/(2)=(1)/(2)((ML^(2))/(3))omega^(2)implies omega^(2)=(3g)/(L)` To find tension at mid point `T-(M)/(2)g=(M)/(2)(a_(cm))_("lower half")` `T-(M)/(2)g=(M)/(2)(omega^(2)(3L)/(4))` `T-(M)/(2)g=(M)/(2)((3g)/(L)(3L)/(4))` `T=(Mg)/(2)+(9Mg)/(8)=(13Mg)/(8)` Stress at mid point = `(T)/(A) = (13Mg)/(8A)` `S = (13dALg)/(8A)=(13dLg)/(8)` |
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