Question

A uniform rod (total length L) pivots at one-quarter from one end. It is pulled to one side through a very small angle and allowed to oscillate in a vertical plane. Let the mass m of the rod equal to 1 kg. (a) Determine the period of oscillation T of the physical pendulum if the total length is 2.06 m (4 points). (b) What is the length of a simple pendulum l that has the same period T as found for the physical pendulum in 11(a) above (3 points)? (c) What is the torsion constant (k) if this pendulum has the same period (T) and inertia (I) as found for the physical pendulum in 11(a) above (3 points)?

Answer 1

b) siueple pendulam will be Is me? /2 madkg L=2.06 2 = dat 14 solution moment of inertial of rot A me a) time period of physiceel pendulaeus *** Z me? 48 T ر1 * LX 12.06]2 0.618858 Kgar y 8 To ܒܐ nge 0.618858 x 4 itxo.8 X 2.06 distance of hinged from con 1s 2T Те 2.27 sec é T? at son e 22 ET 27 / 9:8 I E0:1226189 908 leusth l=11-2016 m c) for torsion constant (K). time pesiod T = 21 E 2.2 = 2T 0.618858 K ? 0.618858 0:12259 . 5.048