Question

A rod of length 1.5 m is attached to a spring of spring constant, k as shown in Fig. II-1. The rod is released from rest when 0 = 0°. While reaching the position 0 = 90°, the angular velocity of the rod was 2.32 rad/s. If the mass of the rod is m kg, determine the value of the spring constant, k. Assume that the spring is unstretched when 0–0°. Take the mass of the rod (m) as 49kg 2 m el 1.5 m A Figure II-1

Answer 1

3 K L=1.5m ng energy In the situo cation mentioned rod is attached at the point o so, the kinatic in the question, the rotational ene of the rod will be due to Circulats motion. on to of the rod due to Torque the applied force. if L the angular and ? be the torque applied then the must have generated with be angular momentum of the red on the rod 2- d L(+) dt d2001 dtz where, Oct) in the angular diaple coment of the rod. I given inertia of of the red with respect to Now, I: moment the point '. we have to find I. mass of the rod Length m arz, of the rod, I=1.5m mana /Length m L L So, moment of Inertia about the point o', [cshare are consider I m L + *** [5 ſm Lan elemental length drat a distamar from the point o L p3 1 m L UT, 1 ) 3 m12

on the Now, we have to look at the applied force of rod. and find out the toroque due to them about point'o' Two forces are applied on the rod. 0 gravitational force (mg) at the center of mass 'm' of the rod which is at (%) distance from o point. 2 Ö, gi So, x=LO @ Force due to the spring of spring constant te at the poin A which is at l distance away from Now, the a torque due to gravity Force x distance. mga Y coso gamg/20 = mg ½ [aso ismall Torquee due to spring force, Force = -kx= - klo Now, xis the linear distance of of spring Torque TE= Forex distance = - kloxl= - Klo So, equeation of motion, mg/-kl2O =- KL Let, Ø=- mg 0--moto w do = do I d20 - -KL2¢ atz :. Circular frequency m1² Now, as, 49kg., and w= 2.32 ro 2.32 rad's Салат [as I = 5 m 12] 2.32 x 49 W/m 88N/m So, the ve value of spring constant, 88N/m . (Aus.) E-KL² Ema l mg to dzo dt² 1 . K22 W= KL2 I Зk m ma 2 sos k= wzm 3 3