Center of Mass

1. Center of Mass Physics Montwood High School R. Casao

2. The center of mass is the point at which all of the mass of an object or system may be considered to be concentrated, for the purposes of linear or translational motion only. • Newton’s second law for the motion of the center of mass: Fnet = m∙acm • The momentum of the center of mass does not change if there are no external forces on the system.

3. The center of mass of a flat object can be found by suspension.

4. The center of mass may be located outside a solid object.

5. The principle of conservation of momentum can be restated in terms of the center of mass. • For several particles of mass m1, m2, etc., let the coordinates of the center of mass of m1 be (x1, y1), those of m2 be (x2, y2), etc. • We define the center of mass of the system as the point having coordinates (xcm, ycm) given by:

6. When a homogeneous body has a geometric center, such as a billiard ball or a sugar cube, the center of mass is at the geometric center. • When a body has an axis of symmetry, such as a wheel or a pulley, the center of mass lies on that axis. • There is no law that says the center of mass has to be within the body. See the donut!

7. When the center of mass of a particle moves’ the x and y components of velocity of the center of mass, vcm-x and vcm-y, is the time derivative of xcm and ycm. Keep in mind velocity = dx/dt.

8. If vx and vy are used to determine the resultant vector v; • Expressing m1 + m2 + m3 + … as M and multiplying both sides by M gives us: • The total momentum is equal to the total mass times the velocity of the center of mass.

9. For a system of particles on which the net external force is 0 N, the total momentum is constant and the velocity of the center of mass is constant. • If the net external force on a system of particles in not 0 N, then total momentum is not conserved and the velocity of the center of mass will change.

10. Taking the time derivative of the velocity,we get acceleration, acm = dvcm/dt. • When a body or a collection of particles is acted on by external forces, the center of mass moves just as tho all the mass were concentrated at that point and it were acted on by a net force equal to the sum of the external forces on the system.

11. Suppose a cannon shell traveling in a parabolic trajectory (neglecting air friction) explodes in flight, splitting into two fragments of equal mass. • The fragments follow new parabolic paths, but the center of mass continues on the original parabolic path as if all the mass were still concentrated at that point.

12. One more useful way of describing the motion of a system of particles:

13. Practice Problem 1 • Three odd-shaped blocks have the following masses and center-of-mass coordinates: (1) 0.3 kg, (0.2 m, 0.3 m); (2) 0.4 kg, (0.1 m, -0.4 m); (3) 0.2 kg, (-0.3 m, 0.6 m). Find the coordinates of the center of mass of the system of three blocks.

14. Center of mass coordinates: x = 0.044 m, y = 0.0556 m.

15. Practice Problem 2 • A 1200 kg station wagon is moving along a straight highway at 12 m/s. Another car of mass 1800 kg and speed 20 m/s has its center of mass 40 m ahead of the center of mass of the station wagon. • A. Find the center of mass of the system consisting of the two automobiles. • Since the cars are on a horizontal surface, you can ignore the y component. The center of mass of the two auto system will lie entirely on the x axis. • Center of mass lies 24 m in front of the center of mass of the station wagon.

16. B. Find the magnitude of the total momentum of the system from the given data. p = 1200kg∙12 m/s + 1800 kg·20 m/s = 14400 kg·m/s + 36000 kg∙m/s p = 50400 kg∙m/s • C. Find the speed of the center of mass of the system.

17. D. Find the total momentum of the system, using the speed of the center of mass. Compare this value to the answer in part B. P= (1200 kg + 1800 kg)∙16.8 m/s = 50400 kg·m/s • The two calculations for the momentum of the system, either individually or using the center of mass of the system, return the same value.