A thin uniformmetallic rod of length0.5 m andradius 0.1 m rotates with an angular velocity 400 rad/s in a horizontal plane about a vertical axis passsing throughone of its ends. Calculate tension in the rod and the elongation of the rod. The density of materialof the rod is 10^(4)kg//m^(3) and Young.s modulus is 2xx10^(11)N //m^(2).

A thin uniformmetallic rod of length0.5 m andradius 0.1 m rotates with

1.	A thin uniformmetallic rod of length0.5 m andradius 0.1 m rotates with an angular velocity 400 rad/s in a horizontal plane about a vertical axis passsing throughone of its ends. Calculate tension in the rod and the elongation of the rod. The density of materialof the rod is 10^(4)kg//m^(3) and Young.s modulus is 2xx10^(11)N //m^(2).
Answer» Solution : (a) Consideran element of lengthdr at a distance .r.from the AXIS of rotation as shown in fig. The CENTRIPETAL forceacting on this ELEMENTWILL be `dT =dm r omega^(2) =(rho A dr) r omega^(2)` As this force is providedby tension in the rod (due to elasticity), so the tension in the rod at a distance from the axis of rotation due to all elementsbetween x = r to x = L. centripetal force due to all elementsbetweenx = r to x = L. i.e., `T=int_(r)^(L) rhoA omega^(2) r dr=(1)/(2) rhoAomega^(2)[L^(2)-r^(2)] ......(1)` So here `T=(1)/(2) XX 10^(4) xx pi xx 10^(-2) xx (400)^(2) [((1)/(2))^(2) -r^(2)]` `=8pi xx 10^(6) [(1)/(4) -r^(2)]N` `T=(1)/(2) rho A omega^(2) (L^(2)-r^(2))` elongationof small element `de =(rho A omega^(2))/( 2Ay) (L^(2)-r^(2)) dr`, Total ELONGATION `e=intde=(rho omega^(2))/(2y) int_(O)^(L)(L^(2)-r^(2))dr=(rho^(2) omega^(2)L^(3))/(3y)`