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Rotation of Rigid Bodies

Date: 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?


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 

Next find the center of mass of the system.

      |<-----0.6--------->|<--c--->|<---0.6-c ----->|   
                                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   
Associated Topics:
High School Physics/Chemistry

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