Quick Review & Centers of Mass Chapter 8.3 March 12, 2007

Let R be the region in the x-y plane bounded by Set up the integral to find the area of the region. [0,2] Bounds: Top Function: Bottom Function: Length: Area:

Set up the integral to find the volume of the solid whose base is the region between the curves: and with cross sections perpendicular to the x-axis equilateral triangles . Length: Area : Volume:

Find the volume of the solid generated by revolving the region defined by about the line y = 3. Length: Outside (R) : Inside (r) : Area : Volume:

Find the volume of the solid generated by revolving the region defined by about the line x = 3. Length: Radius : Area : Volume:

Centers of Mass - Discrete Case Suppose we have two particles of mass m1 and m2 at positions x1 and x2 , and suppose that the particles are connected by a lightweight inflexible rod. We say the particles form a system or a rigid body. Analyzing the properties of such a system is the first step to understanding the properties of more complicated objects.One of the most important properties of the system is the center of mass . Intuitively, the center of mass is the location where an object could balance perfectly level on the tip of a pin.

Centers of Mass - Discrete Case Describe the center of mass for the two particle system where: m1 = m2 m1 < m2 m1 > m2

Toss the bar bell: Center of mass follows basically a parabolic curve where as the ends will rotate around the center of mass.

Sometimes a kicked football sails through the air without rotating, and at other times it tumbles end over end as it travels. With respect to the center of mass of the ball, how is it kicked in both cases? For the tumbling situation, the football is kicked with a torque about its center of mass, so the force has a considerable rotational effect…….

Sometimes a kicked football sails through the air without rotating, and at other times it tumbles end over end as it travels. With respect to the center of mass of the ball, how is it kicked in both cases? For the non-rotating case, the football is kicked so the force is directed at the center of mass, so there is no rotational effect!

See Saw (or Teeter Totter) Each mass exerts a downward force (Force = magnitude of mass x acceleration of gravity)… Whether sitting……

See Saw (or Teeter Totter) Or Standing

See Saw (or Teeter Totter) Or Hanging

TORQUE Each of these forces has a tendency to turn the axis about the origin - this turning effect is call TORQUE and is measured by multiplying the force by the signed distance x from the origin.

Measuring TORQUE Masses to the left of the origin exert negative (counter clockwise) torque Masses to the right of the origin exert positive (clockwise) torque Our system will balance when the |negative torque| = positive torque

See Saw (or Teeter Totter) Our teeter totter will balance if Weight * distance = weight * distance

See Saw (or Teeter Totter) Two children want to balance the see saw. On is 45 pounds and sits 4 ft from the center. Where should the other child sit if he weighs 60 pounds? Weight * distance = weight * distance 45lbs * 4 ft = 60lbs * distance 3 ft = distance

Measuring TORQUE The sum of the torques, measure the tendency of a system to rotate about the origin and is called the system torque. The system will balance if and only if the system torque = 0. Note: 45lbs * (-4) ft + 60lbs * 3 ft = 0 and our teeter totter balanced!

System Torque Let’s find the center of mass for a two particle system between [x1,x2] with m1 at x1 and m2 at x2. Letting represent the center of mass, we measure the distance from the center of mass.

Measuring from Solving for , we get:

We can extend this to any number of particles: Notice the gravity constant cancelled out. That’s because gravity of a feature of the environment, whereas the System Moment is a feature of the system and will remain consistent where ever the system is placed…

Three bodies of mass, 4 kg, 9 kg & 7 kg are located at x1 = -3, x2 = 6, and x3 = 10. If the distances are measured in meters, find the center of mass.

Center of Mass: 2-Dimensional Discrete Case Suppose we have a finite collection of masses in a plane. Each mass mk is located at point (xk,yk). The system mass is: Each mass has a moment about EACH axis: mk’s moment about the x-axis is given by mkyk (yk represents the VERTICAL distance from the x-axis mk’s moment about the y-axis is given by mkxk (xk represents the HORIZONTAL distance from the y-axis)

Center of Mass: 2-Dimensional Discrete Case Suppose we have a finite collection of masses in a plane. Each mass mk is located at point (xk,yk). The system moments are: Moment about the x-axis is given by (yk represents the VERTICAL distance from the x-axis Moment about the y-axis is given (xk represents the HORIZONTAL distance from the y-axis)

Center of Mass: 2-Dimensional Discrete Case Suppose we have a finite collection of masses in a plane. Each mass mk is located at point (xk,yk). The System’s Center of Mass is defined to be:

Particles of masses 2, 5, 8 are located at (-1,3), (0,7) & (2,2) respectively. Find My = 2(-1) + 5(0) + 8(2) = 14 Mx = 2(3) + 5(7) + 8(2) = 57

Find the centroid (center of mass with uniform density) of the region shown, by locating the centers of the rectangles and treating them as point masses…..

Find the centroid (center of mass with uniform density) of the region shown, by locating the centers of the rectangles and treating them as point masses…..

Find the centroid (center of mass with uniform density) of the region shown, by locating the centers of the rectangles and treating them as point masses…..