Answer:
To find the moment of inertia of each link about the center of gravity (CG), we can use the parallel axis theorem, which states that the moment of inertia about an axis parallel to an axis through the center of mass is equal to the moment of inertia through the center of mass plus the product of the mass and the square of the distance between the two axes.
Given the lengths and masses of each link, we can calculate the moment of inertia for each link about its center of gravity.
Link 2:
Length (a) = 4 in
Mass = 0.005394167 blob
To find the moment of inertia about the CG of link 2 (I2), we need to know the distance between the CG and the center of mass. Assuming the center of mass is located at the midpoint of link 2 (which is common for uniform slender links), the distance between the CG and the center of mass is half the length of link 2.
Distance (d2) = a/2 = 4 in / 2 = 2 in
Using the parallel axis theorem, the moment of inertia of link 2 about the CG (I2_CG) is given by:
I2_CG = I2 + m2 * d2^2
where I2 is the moment of inertia about the center of mass, m2 is the mass of link 2, and d2 is the distance between the CG and the center of mass.
Moment of inertia of link 2 about the CG (I2_CG) = I2 + (0.005394167 blob) * (2 in)^2
Link 3:
Length (b) = 5 in
Mass = 0.009494358 blob
Using the same approach as above, the distance between the CG and the center of mass of link 3 (d3) is:
d3 = b/2 = 5 in / 2 = 2.5 in
Moment of inertia of link 3 about the CG (I3_CG) = I3 + (0.009494358 blob) * (2.5 in)^2
Link 4:
Length (c) = 5 in
Mass = 0.012549064 blob
Similarly, the distance between the CG and the center of mass of link 4 (d4) is:
d4 = c/2 = 5 in / 2 = 2.5 in
Moment of inertia of link 4 about the CG (I4_CG) = I4 + (0.012549064 blob) * (2.5 in)^2
Please note that the blob-in^2 is a non-standard unit, and it is unclear how it relates to a standard unit of moment of inertia.