Center of Mass. A baseball thrown into the air follows a smooth parabolic path. A baseball bat thrown into the air does not follow a smooth path.

Center of Mass A baseball thrown into the air follows a smooth parabolic path. A baseball bat thrown into the air does not follow a smooth path. The bat wobbles about a special point. This point stays on a parabolic path, even though the rest of the bat does not. The motion of the bat is the sum of two motions: a spin around this point a movement through the air as if all the mass were concentrated at this point. This point, called the center of mass, is where all the mass of an object can be considered to be concentrated. 1

The motion of the wrench is a combination of straight line motion of its center of mass and rotation around its center of mass. If the wrench were tossed into the air, its center of mass would follow a smooth parabola. 2

Falling Cat The motion of the cat is a combination of straight line motion of its center of mass and rotation around its center of mass. 3

Location of the Center of Mass For a symmetrical object, such as a baseball, the center of mass is at the geometric center of the object. For an irregularly shaped object, such as a baseball bat, the center of mass is toward the heavier end. Examples The center of mass for each object is shown by the red dot. Objects not made of the same material throughout may have the center of mass quite far from the geometric center. 4

Under the action of a constant force of gravity, a body suspended or balanced at its center of mass will be balanced. Only the center of mass of an object needs to be supported in order to support the object. The bird's center of mass is at its beak. 5

The center of mass for this object is under the horse's hind legs. 6

http://www.csiro.au/helix/sciencemail/activities/balchal.html 7

http://library.thinkquest.org/26455/amuse/tricks/fork.shtml 9

Fun with the Center of Mass 11

Static Torque A torque is the rotational counterpart of force. A force tends to change the motion of objects; torque tends to twist or change the rotation of objects. 14

Every time you turn on a water faucet, or tighten a nut with a wrench, you exert a turning force. Torque is produced by this turning force. Torque is different from force. Forces tend to make things accelerate. Torques produce rotation. 15

A torque is produced when a force is applied with leverage. You use leverage when you use a claw hammer to pull a nail out of a piece of wood. The longer the handle of the hammer, the greater the leverage and the easier the task. The longer handle of a crowbar provides even more leverage. WEB Simulation "Wrench" 16

Although the magnitudes of the applied forces are the same in each case, the torque in each case is different. 17

A torque is used when opening a door. A doorknob is placed far away from the turning axis at its hinges to provide more leverage when you push or pull on the doorknob. The direction of your applied force is important. In opening a door, you push perpendicular to the plane of the door. A perpendicular push or pull gives more rotation for less effort. Sample Problem The diagram below shows four forces acting on a door. Which force(s) will cause the door to rotate? F 3 F 4 Hinged Side F 2 F 1 18

Balanced Torques Children can balance a seesaw even when their weights are not equal. Weight alone does not produce rotation torque does. Example A pair of torques can balance each other. Balance is achieved if the torque that tends to produce clockwise rotation by the boy equals the torque that tends to produce counterclockwise rotation by the girl. 19

When a perpendicular force is applied, the lever arm is the distance between the doorknob and the edge with the hinges. If the force is not at right angle to the lever arm, then only the perpendicular component of the force will contribute to the torque. 20

Torque can be calculated using the following equation. r * torque * this symbol represents the Greek letter tau the distance from pivot point to the point of contact; the lever arm F force θ angle between vectors r and F 21

Torque is a vector. The direction of torque is based on the direction in which the force would cause the object to rotate if it were acting alone. CW: clockwise ( ) CCW: counter clockwise (+) 22

YouTube Video 23

Back to the DOOR PROBLEM We can verify our previous answers using the equation below. F 3 F 4 F 2 θ F 1 r F 1 : θ = 0 o sin 0 o = 0 τ = 0 Nm F 3 : r = 0 m τ = 0 Nm F 2 : θ = 180 o sin 180 o = 0 τ = 0 Nm F 4 : r 0 m and sin θ 0 F 4 will cause the door to rotate! 25

Sample Problems - Torque 1. A 4.90 x 10 2 N man stands at the end of a diving board at a distance of 1.5 m from the point at which it is attached to the tower. What is the torque the man exerts on the board? 26

2. A 5.0 kg mass is attached as shown to a pulley of radius 5.0 cm. What torque is produced by the mass? 27

3. A 64 kg painter is standing three fourths of the distance up a ladder that is 3.0 m long. If the ladder makes an angle of 69 o with the ground, what torque does the painter's weight exert on the ladder? 69 o 28

Rotational Equilibrium An important concept that keeps objects, such as buildings, from falling down is rotational equilibrium. An object is in rotational equilibrium if: 1. it has zero velocity v = 0 m/s 2. the net force acting on the object is zero F net = 0 N 3. the net torque acting on the object is zero τ net = 0 Nm 29

What is the net force of acting on the board? Is the board in a state of rotational equilibrium? 10 N F N 10 N Pivot Point What is the net force of acting on the board? Is the board in a state of rotational equilibrium? 10 N Pivot Point 10 N The arrangement of these forces will certainly cause the board to turn. 30

Solving Rotational Equilibrium Problems 1. Draw a picture for the object of interest. Show the relative positions of all the forces and mark any distances. 2. Write an F net equation. 3. Choose a pivot point. 4. Write a τ net equation based on the pivot point. 5. Solve the equations. 31

Net Torque Two forces act on a cylinder as shown in the diagram below. If F 1 = 10.0 N and F 2 = 15.0 N, what is the net torque on the cylinder? F 1 F 2. 5.0 cm 8.0 cm r 1 r 2 32

Net Torque Two forces act on a cylinder as shown in the diagram below. If F 1 = 10.0 N and F 2 = 15.0 N, what is the net torque on the cylinder? F 1 F 2. 5.0 cm 8.0 cm r 1 r 2 33

A massless board serves as a seesaw for two giant hamsters as shown below. One hamster has a mass of 3.0 x 10 kg and sits 2.5 m from the fulcrum. At what distance from the pivot must a 25 kg hamster place himself to balance the seesaw. PIVOT POINT 2.5 m r 2 The seesaw has three forces acting on it, two weights and a normal force at the pivot point. PP 2.5 m r 2 34

There are no forces acting in the x direction therefore, F netx = 0 N. Write a net force equation for all the forces acting in the y direction. F nety = F N W H1 W H2 0 = F N W H1 W H2 This equation will allow us to determine the normal force, but not the position of the second hamster... 35

5. A massless board serves as a seesaw for two giant hamsters as shown below. One hamster has a mass of 3.0 x 10 kg and sits 2.5 m from the fulcrum. At what distance from the pivot must a 25 kg hamster place himself to balance the seesaw. PIVOT POINT 10 2.5 m r 2 36

5. A massless board serves as a seesaw for two giant hamsters as shown below. One hamster has a mass of 30 kg and sits 2.5 m from the fulcrum. At what distance from the pivot must a 25 kg hamster place himself to balance the seesaw. PIVOT POINT 2.5 m r 2 37

We need to write a net torque equation to determine the position of the second hamster. Remember that torque is simply the perpendicular force multiplied by the distance from the pivot point. the torque from Hamster 2 is negative since it will rotate the seesaw clockwise Solve for r 2. 38

Note It is usually a good idea to select the pivot point at the location where an unknown force acts. What do we do if a solid object such as the board in the previous problem has mass? We treat it as if all its mass is concentrated at one point which is called the center of mass. 39

Sample Problem A uniform 1.500 x 10 3 kg beam, 20.0 m long, supports a 1.500 x 10 4 kg box of hamsters 5.00 m from the right support column. Calculate the force on each of the vertical support columns. 20.0 m CM 5.00 m 40

Textbook Questions Page 501, Question 31: Diver Page 529, Question 27: Mountain Climbers Page 501, Question 33: Light Fixture Page 529, Question 28: Crane Problem 43

Sample Problem 46

Sample Problem 47

Sample Problem A uniform beam of mass 50.0 kg and length 3.00 m is attached to a wall with a hinge. The beam supports a sign of mass 3.00 x 10 2 kg which is suspended from its end. The beam is also supported by a wire that makes an angle of 25.0 o with the beam. Determine the components of the force that the hinge exerts and the tension of the wire. 48