17.8: Appendix 2 Homework Problems

Exercise \(\PageIndex{1}\)

A shape is bounded on the left by the \(y\)-axis, on the bottom by the \(x\)-axis, and along its remaining side by the function \(y = - \dfrac{1}{2} x^2 + 8\). Determine the \(x\) and \(y\) coordinates of the centroid of this shape via integration. (Hint: for \(\bar{y}\), work from the top down to make the math easier.)

Figure \(\PageIndex{1}\): problem diagram for Exercise \(\PageIndex{1}\). A shape in the first quadrant of a Cartesian coordinate plane is bounded by the intersection of the function \(y = - \dfrac{1}{2} x^2 + 8\) with the \(x\)- and \(y\)-axes.

Solution:

\(\bar{x} = 1.5 \ cm, \, \bar{y} = 3.2 \ cm\)

Exercise \(\PageIndex{2}\)

Determine the \(x\) and \(y\) coordinates of the centroid of the shape shown below via integration.

Figure \(\PageIndex{2}\): problem diagram for Exercise \(\PageIndex{2}\). A trapezoid lies in the first quadrant of the Cartesian coordinate plane, with two sides lying along the axes.

Solution:

\(\bar{x} = 2.94 \ in, \, \bar{y} = 4.24 \ in.\)

Exercise \(\PageIndex{3}\)

A water tank as shown below takes the form of an inverted, truncated cone. The diameter of the base is 4 ft, the diameter of the top is 8 ft, and the height of the tank is 4 ft. Using integration, determine the height of the center of mass of the filled tank. (Assume the tank is filled with water and the walls have negligible mass.)

Figure \(\PageIndex{3}\): problem diagram for Exercise \(\PageIndex{3}\). A water tank in the shape of an inverted truncated circular cone, with a wide top and a narrower base.

Solution:

\(z_c = 2.43 \ ft\)

Exercise \(\PageIndex{4}\)

Use the method of composite parts to determine the centroid of the shape shown below.

Figure \(\PageIndex{4}\): problem diagram for Exercise \(\PageIndex{4}\). A composite shape in the first quadrant of a Cartesian coordinate plane, with two sides lying along the axes, consists of two trapezoids, or two rectangles and two right triangles.

Solution:

\(x_c = 1.14 \ in, \, y_c = 1.39 \ in\)

Exercise \(\PageIndex{5}\)

A floating platform consists of a square piece of plywood weighing 50 lbs with a negligible thickness on top of a rectangular prism of a foam material weighing 100 lbs as shown below. Based on this information, what is the location of the center of mass for the floating platform?

Figure \(\PageIndex{5}\): problem diagram for Exercise \(\PageIndex{5}\). A foam rectangular prism with one half of its top face covered by a plywood square lies in the first octant of a Cartesian coordinate system.

Solution:

\(x_c = 3.33 \ ft, \, y_c = 1.33 \ ft, \, z_c = 2 \ ft\)

Exercise \(\PageIndex{6}\)

Use the integration method to find the moments of inertia for the shape shown below…

• About the \(x\)-axis through the centroid.
• About the \(y\)-axis through the centroid.

Figure \(\PageIndex{6}\): problem diagram for Exercise \(\PageIndex{6}\). An downwards-facing isosceles triangle has a horizontal base and a center of mass on its line of symmetry.

Solution:

\(I_{xx} = 6.075 * 10^{-7} \ m^4\)

\(I_{yy} = 5.0625 * 10^{-8} \ m^4\)

Exercise \(\PageIndex{7}\)

Use the integration method to find the polar moment of inertia for the semicircle shown below about point O.

Figure \(\PageIndex{7}\): problem diagram for Exercise \(\PageIndex{7}\). A semicircle lies with its straight edge centered on the \(x\)-axis of a Cartesian coordinate plane, with origin O.

Solution:

\(J_{zz} = 1017.9 \ in^4\)

Exercise \(\PageIndex{8}\)

A plastic beam has a square cross-section with semicircular cutouts on the top and bottom as shown below. What is the area moment of inertia of the beam’s cross section about the \(x\) and \(y\) axes through the center point?

Figure \(\PageIndex{8}\): problem diagram for Exercise \(\PageIndex{8}\). A beam cross-section consists of a square with semicircular cutouts centered on the top and bottom sides.

Solution:

\(I_{xx} = 7.08 \ in^4, \, I_{yy} = 17.36 \ in^4\)

Exercise \(\PageIndex{9}\)

A piece of angled steel has a cross section that is 1 cm thick and has a length of 6 cm on each side as shown below. What are the \(x\) and \(y\) area moments of inertia through the centroid of the cross section?

Figure \(\PageIndex{9}\): problem diagram for Exercise \(\PageIndex{9}\). A part cross-section consists of an L shape composed of two identically sized rectangles.

Solution:

\(I_{xx} = I_{yy} = 35.462 \ cm^4 = 3.546 * 10^{-7} \ m^4\)

Exercise \(\PageIndex{10}\)

The pendulum in an antique clock consists of a brass disc with a mass of 0.25 kg and diameter of 6 cm at the end of a slender wooden rod with a mass of 0.1 kg. Determine the mass moment of inertia of the pendulum about the top of the rod.

Figure \(\PageIndex{10}\): problem diagram for Exercise \(\PageIndex{10}\). A clock pendulum is represented as a vertical wooden rod with a brass disc attached to its bottom edge.

Solution:

\(I_{zz} = 0.02026 \ kg \ m^2\)

Exercise \(\PageIndex{11}\)

A space telescope can be approximated as a 600-kg cylinder with a 4-meter diameter and 4-meter height attached to two 100-kg solar panels as shown below. What is the approximate mass moment of inertia for the space telescope about the \(y\)-axis shown?

Figure \(\PageIndex{11}\): problem diagram for Exercise \(\PageIndex{11}\). A space telescope is represented as a vertical cylinder with two horizontal rods extending from midway along the cylinder's height, each supporting a solar panel.

Solution:

\(I_{yy} = 10,216.7 \ kg \ m^2\)

Exercise \(\PageIndex{12}\)

A flywheel has an original weight of 15 pounds and a diameter of 6 inches. To reduce the weight, four two-inch diameter holes are drilled into the flywheel, each leaving half an inch to the outside edge as shown below. What was the original polar mass moment of about the center point? Assuming a uniform thickness, what is the new mass moment of inertia after drilling in the holes? (Hint: holes count as negative mass in the mass moment calculations.)

Figure \(\PageIndex{12}\): problem diagram for Exercise \(\PageIndex{12}\). A flywheel consists of a circular disk with four circular holes drilled in a radially symmetric pattern about its center point.

Solution:

\(J_{without holes} = 0.01456 \ slug \ ft^2\)

\(J_{with holes} = 0.01060 \ slug \ ft^2\)