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Question

Need complete explanation withSteps to solve.  5 Stars for best explanation

A 3-D assembly: Now consider a 10 kg rectangular bar and a 5 kg circular cylinder, with dimensions shown in meters, that are attached together to form a rigid body. The bar is centered on the top surface of the cylinder. All dimensions are meters. Find the CG and then find the moment of inertia of the assembly about the y-axis through the CG. Next find the moment of inertia about the x-axis through the CG. Finally, compute the moment of inertia of the assembly about a y-axis through the center of the top surface of the rectangular bar. You will need to use the parallel axis theorem along with the results from previous parts of this example. The CG should be 0.8333 m from the bottom of the disk. The (I G) y should be 5.433 kg-m2. Using a y-axis through the center of the top, the result for (I top) y should be 22.50 kg-m2.

Explanation / Answer

Take bottom centre of disk as x,y,z = (0,0,0).

x_cg = ((5*0.4/2 + 10*(0.4 + 1.5/2)) / (5 + 10)

x_cg = 0.8333 m

Iy_disk (about its own c.g) = m(3R^2 + h^2) /12 = 5*(3*0.6^2 + 0.4^2) /12 = 0.5167 kg-m^2

Iy_disk about c.g of assembly = 0.5167 + 5*(0.833 - 0.4/2)^2 = 2.522 kg-m^2

Iy_bar (about its own c.g) = m/12*(0.2^2 + 1.5^2) = 10/12 * (0.2^2 + 1.5^2) = 1.90833 kg-m^2

Iy_bar about c.g of assembly = 1.90833 + 10*(0.4 + 1.5/2 - 0.8333)^2 = 2.911 kg-m^2

Total Iy_assembly = 2.522 + 2.911 = 5.433 kg-m^2

b)

By symmetry, Ix = Iy = 5.433 kg-m^2

c)

Iy_top = 5.433 + (10+5)*(0.4 + 1.5 - 0.833333)^2 = 22.5 kg-m^2

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