Title: Center of Mass Concept
1Center of Mass Concept
- The center of mass is also referred to as the
center of gravity. We can locate the center of
mass (CM) in 2 different ways shown in the next 2
slides
2Locating Center of Gravity 1
- The center of gravity can be thought of as a
point within a body where all of the weight of
the body is located. Thus we can locate this
point by balancing the object on a fulcrum, such
as on the seesaw shown here
3Locating Center of Gravity 2
- The center of gravity will always move to a
stable equilibrium point directly below a point
of suspension. Thus we can use the technique
illustrated in this diagram to experimentally
locate the CM point
4Calculating the CM point
5Calculating the CM point 2
See Active Figure 9.17 at pse6.com
6Calculating CM point in continuous objects
7Motion of the CM point
The CM point follows the normal parabolic
trajectory of any point object. However, the
other points within the object simultaneously rota
te around the CM point. See Active Figure 9.16
at pse6.com
8Moment of Inertia of Point Objects
Unit kg-m2
9Example of Moment of Inertia for Point Objects
I 2 (mb2 Ma2)
10Calculation of Moment of Inertia for Continuous
Object
- dm ldx for straight line
- dm lRdq for thin ring (hoop)
- l units kg/m
11Moment of Inertia for straight line object
12Moment of Inertia for straight rod about an axis
through CM
13Moment of Inertia for straight rod about an axis
through one end
14Moment of Inertia of Thin Ring (Hoop)
15(No Transcript)
16Moment of Inertia of Cylinder/Disk
17Moment of Inertia of Solid Sphere
See Active Figure 10.30 at pse6.com
18Derivation of Parallel Axis Thm.
19Example of Parallel Axis Thm.
- Given ICM for straight rod (1/12)ML2,
- use the Parallel Axis Thm. to calculate
- the moment of inertia for an axis through
- one end of the rod
- Solution
- I ICM MD2 (1/12)ML2 M(L/2)2
-
- I (1/12 ¼)ML2 (1/3)ML2
- Note that above result equals the result we get
when directly integrating.