1. A rigid body of total mass M and total length L consists of two solid uniform cylinders joined rigidly together at their end faces, as shown, with their principal axes of symmetry lying along the x-axis.
See attached file.
Let thinner and thicker parts have masses m1 and m2 respectively and volumes v1 and v2 respectively.
v1 = (L/2)*pi*(2a)^2 = 2*pi*L*a^2
v2 = (L/2)*pi*a^2 = (1/4)v1
v1:v2 = 1:4
m1 = M*1/(1+4) = M/5
m2 = 4M/5
Ix = (1/2)*(M/5)*a^2 + (1/2)*(4M/5)*(2a)^2
=> Ix = 17*M*a^2/10 --Answer
Iy = I1c + m1*x1c^2 + I2c + m2*x2c^2 + Ix/2
where, I1c and I2c are the moment of inertias of thinner and thicker part about their centroid assuming them thin rods. x1c and x2c are the distances of centroid of each part from the y axis. Ix/2 is added because of perpendicular theorem which says
for 2-dimensional objects like thin disk:
Ix = Iy + Iz = 2*Iy (in case of symmetric circular ...
The solution shows all the workings to arrive at the answers to the problems. The inertia of objects, cable tension, force on a hinge and a torque.