Dynamics of a four-bar linkageI = 4.27x10 kg -m²M = 0.08kgM, = 0.26kgM = 0.18kgOA =0.08mAB = 0.26mI, =1.46 x 10 'kg-m²ANВС - 0.18mI = 4.86 x10 kg m²%3DOC = 0.18mg = 9.8m/s +BT=0.1N mhow was I(inertia) calculated? what is the formula?

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Dynamics of a four-bar linkage I, = 4.27x10ʻkg-m² I, =1.46 x10 °kg-m² I, = 4.86x10“kg-m² OA=0.08m M = 0.08kg AB = 0.26m M, = 0.26kg BC = 0.18m M =0.18kg OC = 0.18m g = 9.8m/s² 4 B T=0.IN·m how was l(inertia) calculated? what is the formula?

Dynamics of a four-bar linkage I, = 4.27x10ʻkg-m² I, =1.46 x10 °kg-m² I, = 4.86x10“kg-m² OA=0.08m M = 0.08kg AB = 0.26m M, = 0.26kg BC = 0.18m M =0.18kg OC = 0.18m g = 9.8m/s² 4 B T=0.IN·m how was l(inertia) calculated? what is the formula?

Elements Of Electromagnetics

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Author:Sadiku, Matthew N. O.

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Question

Dynamics of a four-bar linkage
I = 4.27x10 kg -m²
M = 0.08kg
M, = 0.26kg
M = 0.18kg
OA =0.08m
AB = 0.26m
I, =1.46 x 10 'kg-m²
AN
ВС - 0.18m
I = 4.86 x10 kg m²
%3D
OC = 0.18m
g = 9.8m/s +
B
T=0.1N m
how was I(inertia) calculated? what is the formula?

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