Asked by Lilly
A 20 kg square slab with side lengths 3.0 m rotates around an axis perpendicular to the slab. A particle of mass 0.60 kg sits on each of the corners of the square. Also, a particle of mass 0.80 kg sits at the center of each of the edges. What is the moment of inertia of the system?
Help me please :)
Help me please :)
Answers
Answered by
Elena
m1=20 kg, a=3 m,
I1 =m•a²/6 = 20•9/6 =30 kg•m²
m2 =0.6 kg , b =a•√2/2=3•√2/2 =2.13 m.
I2 =mb²=0.6•2.13²= 2.7 kg•m²
m3 = 0.8 kg, c=a/2 =1.5 m.
I3=mc²=0.8• 1.5² =1.8 kg•m².
I =I1+4•I2+4•I3 = 30+4•(2.7+1.8) =48 kg•m².
I1 =m•a²/6 = 20•9/6 =30 kg•m²
m2 =0.6 kg , b =a•√2/2=3•√2/2 =2.13 m.
I2 =mb²=0.6•2.13²= 2.7 kg•m²
m3 = 0.8 kg, c=a/2 =1.5 m.
I3=mc²=0.8• 1.5² =1.8 kg•m².
I =I1+4•I2+4•I3 = 30+4•(2.7+1.8) =48 kg•m².
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