Determine the coordinates of the mass center of the body con

Determine the coordinates of the mass center of the body constructed of steel plate and a steel rod. The plate thickness is 2 mm. The rod diameter is4 mm and its length is 50 mm. The origin O is at the mid-thickness of the plate, and the rod is welded to the face of the plate. What is the mass of the body? Determine mass moment of inertia with respect to x, y, and z. [Ans. X=1.845mm, Y=0, Z=6.02mm, m=0.0693kg]

The more explanation the better.

Use density of steel is 7830 kg/m^3

1l Q1. Determine the coordinates of the mass center of the body constructed of steel plate and a steel rod. The plate thickness is 2 mm. The rod diameter is4 mm and its length is 50 mm. The origin O is at the mid-thickness of the plate, and the rod is welded to the face of the plate. What is the mass of the body? Determine mass moment of inertia with respect to x, y, and z. Ans. X=1.845mm, Y=0, Z=6.02mm, m=0.0693kg]T 2 40 mm 50 mm 20 mm

Solution

we have the density, so we need first the volume

the volume of the semi plate is:

the volume of the rectangle below= 20x(40+40)x2= 3200 mm^3

the volume of the semi circle= (1/2)xpix(40^2)x2=5026.55 mm^3

so this volume is 3200+5026.55= 8226.55 mm^3

now the volume of the rod is

volume of the rod= 50xpix(4/2)^2 = 628.319 mm^3

now using the denity we can know the mass of the 2 bodys

mass semi circle with rectangle= 7830 (kg/m^3) x (8226.55 mm^3 / 1000^3) = 0.064414 kg

mass of the rod= 7830 x 628.319 / 1000^3 = 0.00492 kg

the total mass is

0.064414+0.00492= 0.069334 kg

know the center of mass is

starting in the point O

using this equation

r=(1/2)(m1r1 +r2m2 ) / (m1+m2) where m1 is the mass of body 1 and m2 is the mass of body 2 and r1 is the center of inertia of one and r2 for 2

FOR x

rx=(1/2) ((25x 0.00492)+(0.064414 x2)/(0.064414+0.00492))=1.81605 mm

FOR y

ry=0

because all the center of inertia of the semi circle, the rectangle and the rod are in hte point O in the center

FOR z

the center of inertia for the rod is 0, so we divided the body 1 in the semi circle and the rectangle to know the center of inertia of all the body

mass of semi circle=5026.55 x (7830/1000^3)= 0.039358 kg

mass of rectangle=3200 x (7830/1000^3) =0.025056 kg

RCircle=4R/3pi = 4x40/3x3.1416 = 16.9765 mm

ry=(1/2)(0.039358x16.9765 + 0.025056x10) / (0.039358+0.025056) = 7.16168 mm aprox, maybe the decimals but the answer is very close

good luck