Question #108070
Four objects are held in position at the corners of a rectangle by light rods as shown in Figure P8.37. Find the moment of iner- tia of the system about (a)thex-axis,(b)they-axis, and (c) an axis through O and perpendicular to the page
1
Expert's answer
2020-04-06T09:03:56-0400

Solution

The moment of inertia of a mass m placed r meters from the axis of rotation is


I=mr2.I=mr^2.


This is is an additive quantity. Therefore, the moment of inertia around x-axis will be


Ix=m1r12+m2r22+m3r32+m4r42,I_x=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,

where

Ix=m1r12+m2r22+m3r32+m4r42,r1=r2=r3=r4=3 m.Ix=(3.1+1.7+4.5+2.3)32=104 kgm2.I_x=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,\\r_1=r_2=r_3=r_4=3\text{ m}.\\ I_x=(3.1+1.7+4.5+2.3)3^2=104\text{ kg}\cdot\text{m}^2.

Consider rotation around y-axis:


Iy=m1r12+m2r22+m3r32+m4r42,r1=r2=r3=r4=2 m.Iy=(3.1+1.7+4.5+2.3)22=46.4 kgm2.I_y=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,\\r_1=r_2=r_3=r_4=2\text{ m}.\\ I_y=(3.1+1.7+4.5+2.3)2^2=46.4\text{ kg}\cdot\text{m}^2.

Now consider rotation around z-axis:


Iz=m1r12+m2r22+m3r32+m4r42,r1=r2=r3=r4=13 m.Iz=(3.1+1.7+4.5+2.3)132=151 kgm2.I_z=m_1r_1^2+m_2r^2_2+m_3r^2_3+m_4r^2_4,\\r_1=r_2=r_3=r_4=\sqrt{13}\text{ m}.\\ I_z=(3.1+1.7+4.5+2.3)\sqrt{13^2}=151\text{ kg}\cdot\text{m}^2.

Conclusion: the further, the higher.


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