Question

14 (a) A small collar of mass 1,0 kg is attached to one end of a uniform rod of mass, 3,0 kg and length I m Calculate the distance of the centre of gravity of the combination from the other end of the rod, the radius of gyration about the centre of gravity, the period of oscillation as a compound pendulum, suspended from the other end of the rod [0,625 m, 0,33 m, 1,79 ]

Answer 1

go o let point be the refenrence point on on A = 10,0) . 3 kq centre of gravity = may, tmay 0.5m to c 0-8250m 10.375 mitm2 mi= 3kg 4, = centre of mass of distence of my from A - 0.5m m₂ = 1kg - Im. C.O.G= n 3xo.st tx 4 0625m from A. o We know moment of Inertia I= mk² where K= radius of gyration here I is Mor of m, & my about centre of gravity c. I= 1 + 12 I, - mic? + (mo)x fo.125) La mx lo-375) 1 = 3 + + 3x10-125 + 1x0375² 12

2= 0.4375 = (mitm) xk² K = 0.4375 1+3 c). (K = 0.33 Now the differential equation for the torque on the pendulum s 1. ö = -(migoit my gr2.). 8 where r arg are distence of centre of mass mik me from the pivot = 0.5 : 2 12 Mol about pivot = m, L2 + mixlo.s) ² 12 3.41 + 3x0 15" + 12 I = 2 kgm? Bxo-st141)*9.8) Imurit mom?) g wa 3.50 Pa 21. 21 . 1.794 Second 3.50