Center of Mass of a Composite Object The composite object in Fig. 7.36 consists of a bar welded to a cylinder. The homogeneous bar is aluminum (weight density 168 Ib/ft³), and the homogeneous cylinder is bronze (weight density 530 Ib/ft³). Determine the center of mass of the object.

Question

Center of Mass of a Composite Object

The composite object in Fig. 7.36 consists of a bar welded to a cylinder. The homogeneous bar is aluminum (weight density 168 Ib/ft³), and the homogeneous cylinder is bronze (weight density 530 Ib/ft³). Determine the center of mass of the object.

Strategy
We can determine the weight of each homogeneous part by multiplying its volume by its weight density. We also know that the center of mass of each part coincides with the centroid of its volume. The centroid of the cylinder is located at its center, but we must determine the location of the centroid of the bar by treating it as a composite volume.

Accepted Answer

The volume of the cylinder is 12[π(4)² - π(2)²] = 452 in.³ = 0.262 ft³ so
its weight is

[latex]W_{(cylinder)} = (0.262)(530) = 138.8 Ib .[/latex]

The x coordinate of its center of mass is [latex]\overline{x}_{(cylinder)} = 10[/latex] in .

The volume of the bar is [latex](10)(8)(2) + \frac{1}{2} \pi (4)^{2}(2) - \frac{1}{2} \pi (4)^{2}(2) = 160 in .^{3} = 0.0926 ft^{3}[/latex] , and its weight is

[latex]W_{(bar)} = (0.0926) (168) = 15.6 Ib.[/latex]

We can determine the centroid of the volume of the bar by treating it as a composite volume consisting of three parts (Fig. a). Part 3 is a semicircular "cutout." The centroids of part 1 and the semicircular cutout 3 are located at the centroids of their semicircular cross sections (Fig b). Using the information summarized in Table 7.8, we have

Table 7.8 Information for determining the x coordinate of the centroid of the bar

	[latex]\pmb{\overline{x}_{i}(in.)} [/latex]	[latex]\pmb{V_{i} (in^{3} )}[/latex]	[latex] \pmb{\overline{x}_{i}V_{i} (in^{4} )} [/latex]
Part 1	[latex]-\frac{4(4)}{3π} [/latex]	[latex]\frac{1}{2}\pi (4)^{2}(2) [/latex]	[latex]-\frac{4(4)}{3\pi }[\frac{1}{2} \pi (4)^{2}(2) ] [/latex]
Part 2	5	(10)(8)(2)	5[(10)(8)(2)]
Part 3	[latex]10 - \frac{4(4)}{3\pi }[/latex]	[latex] - \frac{1}{2}\pi (4)^{2}(2) [/latex]	[latex]-\left[10 - \frac{4(4)}{3\pi }[\frac{1}{2}\pi (4)^{2} (2) ] \right][/latex]

[latex]\overline{x} _{(bar)} = \frac{\overline{x} _{1}V_{1} + \overline{x} _{2}V_{2} + \overline{x} _{3}V_{3} }{V_{1} + V_{2} + V_{3} } = \frac{ - \frac{4(4)}{3\pi }\left[ \frac{1}{2}\pi (4)^{2} (2) \right] + 5\left[(10)(8)(2)\right] - \left[10 - \frac{4(4)}{3\pi } \right][ {\frac{1}{2}\pi (4)^{2}(2) }] }{\frac{1}{2} \pi (4)^{2}(2) + (10)(8)(2) - \frac{1}{2} \pi (4)^{2}(2) } = 1.86 in . [/latex]

Therefore the x coordinate of the center of mass of the composite object is

[latex]\overline{x} = \frac{\overline{x}_{(bar) }W_{(bar)}+ \overline{x} _{(cylinder)} W_{(cylinder)}}{W_{(bar)} + W_{(cylinder)} } = \frac{(1.86)(15.6) + (10)(138.8) }{15.6 + 138.8} = 9.18 in .[/latex]

Because of the symmetry of the bar, the y and z coordinates of its center of mass are [latex] \overline{y} = 0 and \overline{z} = 0.[/latex]

Question 7.19: Center of Mass of a Composite Object The composite object in...

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