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Transfer Theorem for Moment of Inertia The Transfer Theorem is also known as Parallel-Axis Theorem. When we calculated the area and mass moments of inertia via integration, one of the first things we had to do was to select a point or axis we were going to take the moment of inertia about. We then measured all distances from that point or axis, where the distances were the moment arms in our moment integrals. If we pick a different point or axis to serve as the center all of these distances will be different, which means that we will get a different moment of inertia.

The distances used in our moment integrals depend on the point or axis chosen. For the instance on the right however, each of these distance vectors can be broken down into a vector from the origin to the centroid, and then a vector out from the centroid to all the points in the shape. Similarly, we can find the overall moment of inertia by adding two sets of moment integrals. The first moment integral will add up all the distance vectors from the origin to the centroid (This will be like a point mass on a massless stick, so we get a mass times distance squared term), the second will be the moment integral about the centroid (which is what is listed in the tables). By adding these two terms together we can find the moment of inertia about the given origin point.

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The Parallel Axis Theorem states that a bodies moment of inertia about any given axis is the moment of inertia about the centroid plus the mass of the body times the distance between the point and the centroid squared This works for both mass and area moments of inertia as well as for both rectangular and polar moments of inertia. Above the mass moment of inertia is listed, but if we substitute in areas instead of masses we can use it for area moments of inertia.

For mass moments of inertia we will use the mass of the body and for area moments of inertia we will use the area of the shape.

For rectangular area moments of inertia and for 3D mass moments of inertia the distances in the equation will be the distance between the axis or rotation and the centroid while for 2D polar moments of inertia we will measure the distances from the point of rotation to the centroid.

With these equations we can see that the moment of inertia of a body is always lowest about its centroid (where d = 0), and that the further we move away from the centroid the larger the moment of inertia will become.

The moment of inertia of an area about any axis is defined as the moment of inertia of the area about the parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance d from the centroidal axis to the axis in question.

Example #1: Find the moment of inertia of the dumbbell in Example 8-5 rotated about one end of the dumbbell.

Parallel-axis theorem: I = ICM + Mh2 ICM = 144 kg x m2 h = 1.20 m M = 50.0 kg + 50.0 kg = 100.0 kg I = ICM + Mh2 = 144 kg x m2 + (100.0 kg)(1.20 m)2 = 288 kg x m2

Example #2: A solid sphere with mass 60.0 kg and radius 0.150 m has a moment of inertia for rotation through its central axis. What will be the moment of inertia of the sphere, if the rotation axis is changed to pass through a point on its surface? Answer: If the rotation axis is changed from the center axis of the sphere to an axis that connects to the surface of the sphere, then the distance between these points is the radius of the sphere: d = 0.150 m. The mass and moment of inertia through the sphere's center of mass are given. Using these, the moment of inertia for the parallel axis can be calculated using the formula:

The moment of inertia for rotation around the axis at the surface of the sphere is

Exersices 1. A long rod with mass has a moment of inertia, for rotation around an axis near one end. The total length of the rod is 10.00 m. What is the moment of inertia for rotation through the rod's center?

2. Determine the moment of inertia of the shaded area shown with respect to each of the coordinate axes.

3. Determine the moment of inertia of the shaded area shown with respect to each of the coordinate axes.

4. Determine the moments of inertia of the beamís cross-sectional area shown about the x and y centroidal axes.

5. Determine the moments of inertia and the radius of gyration of the shaded area with respect to the x and y axes.

6. A child’s swing at a rustic amusement park is made up of an old tire hanging from a lightweight rope tied to a tree limb. The tire has a radius of 0.310 m and a mass of 11.0 kg, and hangs from a rope 2.50 m long. If the rope–tire combination is swinging around the tree limb at 1.20 rad>s at the low point of its motion, what is its kinetic energy?

7. Determine the moment of inertia.

8.

[1] [2] Two I shaped sections are welded together. [3]What is the moment of inertia about Axis A-A of this composites sections?

9. A rod of mass M= 3Kg and length pivots about an axis, perpendicular to its length, which passes through one of its ends. What is the moment of inertia of the rod?

10. A beam is made by connecting two 2" x 4" beams in a T pattern with the cross section as shown below. Determine the location of the centroid of this combined cross section and then find the rectangular area moment of inertia about the x axis through the centroid point

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The distances used in our moment integrals depend on the point or axis chosen. For the instance on the right however, each of these distance vectors can be broken down into a vector from the origin to the centroid, and then a vector out from the centroid to all the points in the shape. Similarly, we can find the overall moment of inertia by adding two sets of moment integrals. The first moment integral will add up all the distance vectors from the origin to the centroid (This will be like a point mass on a massless stick, so we get a mass times distance squared term), the second will be the moment integral about the centroid (which is what is listed in the tables). By adding these two terms together we can find the moment of inertia about the given origin point.

.

The Parallel Axis Theorem states that a bodies moment of inertia about any given axis is the moment of inertia about the centroid plus the mass of the body times the distance between the point and the centroid squared This works for both mass and area moments of inertia as well as for both rectangular and polar moments of inertia. Above the mass moment of inertia is listed, but if we substitute in areas instead of masses we can use it for area moments of inertia.

For mass moments of inertia we will use the mass of the body and for area moments of inertia we will use the area of the shape.

For rectangular area moments of inertia and for 3D mass moments of inertia the distances in the equation will be the distance between the axis or rotation and the centroid while for 2D polar moments of inertia we will measure the distances from the point of rotation to the centroid.

With these equations we can see that the moment of inertia of a body is always lowest about its centroid (where d = 0), and that the further we move away from the centroid the larger the moment of inertia will become.

The moment of inertia of an area about any axis is defined as the moment of inertia of the area about the parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance d from the centroidal axis to the axis in question.

Example #1: Find the moment of inertia of the dumbbell in Example 8-5 rotated about one end of the dumbbell.

Parallel-axis theorem: I = ICM + Mh2 ICM = 144 kg x m2 h = 1.20 m M = 50.0 kg + 50.0 kg = 100.0 kg I = ICM + Mh2 = 144 kg x m2 + (100.0 kg)(1.20 m)2 = 288 kg x m2

Example #2: A solid sphere with mass 60.0 kg and radius 0.150 m has a moment of inertia for rotation through its central axis. What will be the moment of inertia of the sphere, if the rotation axis is changed to pass through a point on its surface? Answer: If the rotation axis is changed from the center axis of the sphere to an axis that connects to the surface of the sphere, then the distance between these points is the radius of the sphere: d = 0.150 m. The mass and moment of inertia through the sphere's center of mass are given. Using these, the moment of inertia for the parallel axis can be calculated using the formula:

The moment of inertia for rotation around the axis at the surface of the sphere is

Exersices 1. A long rod with mass has a moment of inertia, for rotation around an axis near one end. The total length of the rod is 10.00 m. What is the moment of inertia for rotation through the rod's center?

2. Determine the moment of inertia of the shaded area shown with respect to each of the coordinate axes.

3. Determine the moment of inertia of the shaded area shown with respect to each of the coordinate axes.

4. Determine the moments of inertia of the beamís cross-sectional area shown about the x and y centroidal axes.

5. Determine the moments of inertia and the radius of gyration of the shaded area with respect to the x and y axes.

6. A child’s swing at a rustic amusement park is made up of an old tire hanging from a lightweight rope tied to a tree limb. The tire has a radius of 0.310 m and a mass of 11.0 kg, and hangs from a rope 2.50 m long. If the rope–tire combination is swinging around the tree limb at 1.20 rad>s at the low point of its motion, what is its kinetic energy?

7. Determine the moment of inertia.

8.

[1] [2] Two I shaped sections are welded together. [3]What is the moment of inertia about Axis A-A of this composites sections?

9. A rod of mass M= 3Kg and length pivots about an axis, perpendicular to its length, which passes through one of its ends. What is the moment of inertia of the rod?

10. A beam is made by connecting two 2" x 4" beams in a T pattern with the cross section as shown below. Determine the location of the centroid of this combined cross section and then find the rectangular area moment of inertia about the x axis through the centroid point

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