Question

20.15 Rotation of a small drum of mass moment of inertia 0.44 kg.m" and with frictional resistance torque 0.3 N.m is initiated by a string wound around its 20 mm diameter shaft and attached to a stretched spring, as shown in Figure 20.7. If the spring, which has a stiffness 2 N/mm, is initially stretched by 120 mm before the drum is released from rest, determine the maximum angular velocity, in revolutions per minute, attained by the drum

please show all lines of working

thanks in advance

Answer 1

llllll 15x2000 I = 0.44 kg-m? V = 0-3 N-m r= 10x1őkm K= 2000 N/m xi= 0.120m Maximam angulaes relocity will be atterned by drum. when Tr= 7 0 K2ex loxlö} = 0.3 = xf = _ = =0.015m Now By work-Kinetic energy theorem. € 42,7 = 4 x 2 + 1 I ² + TEO +(2+6) = { [W2 +0.3 x 92 = 2000.123.016°) - 0.54.44w40.34 W2 2x. %44 + loudl •12?.0152) = -22w2+ .2202= .44602 ?W? 32.22 =W= 5.7 radls = 54.5 freu min de