Rotation of Rigid BodiesDate: 02/05/99 at 06:18:05 From: Anonymous Subject: Rotation of Rigid Bodies In this problem, there is a uniform rod of length 1.2m and mass 0.8kg. Placed at one end is a 0.2 kg mass and at the other a 0.5 kg mass. The angular speed of the rod when rotating in a horizontal plane is 5 rad/s. Find the KE when the rod is spinning on the axis that is through the center of mass of the laden rod. When I have found the center of mass, how do I find the moment of inertia of the system? Thanks! Student Date: 02/05/99 at 16:29:29 From: Doctor Anthony Subject: Re: Rotation of Rigid Bodies Finding the Moment of Inertia of the uniform rod about its center: Length 2a and mass M gives MI = Ma^2/3 = 0.8 x 0.6^2/3 = 0.096 kg.m^2 Now use the parallel axis theorem to get the MI of the rod about the center of mass of the whole system. If c = distance to center of mass, the MI = 0.096 + 0.8c^2 about center. Next find the center of mass of the system. |<-----0.6--------->|<--c--->|<---0.6-c ----->| 0.2------------------0.8-------|---------------0.5 center of mass Taking moments about the center of mass: 0.2(c+0.6) + 0.8c = 0.5(0.6-c) 0.2c + 0.12 + 0.8c = 0.3 - 0.5c 1.5c = 0.18 c = 0.12 So the MI of the rod about this center of mass = 0.096 + 0.8 x 0.12^2 = 0.10752 The MI of the 0.2 mass about the center of mass = 0.2 x 0.72^2 = 0.10368 The MI of the 0.5 mass about the center of mass = 0.5 x 0.48^2 = 0.1152 Add these three answers to get: The MI of whole system about the center of mass = 0.3264 The KE if the rod is spinning at 5 rads/sec about the center of mass is (1/2)I.w^2 = (1/2)x 0.3264 x 25 = 4.08 N.m - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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