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Physics CHAPTER 8 ROTATIONAL MOTION. The Radian  The radian is a unit of angular measure  The radian can be defined as the arc length s along a circle.

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Presentation on theme: "Physics CHAPTER 8 ROTATIONAL MOTION. The Radian  The radian is a unit of angular measure  The radian can be defined as the arc length s along a circle."— Presentation transcript:

1 Physics CHAPTER 8 ROTATIONAL MOTION

2 The Radian  The radian is a unit of angular measure  The radian can be defined as the arc length s along a circle divided by the radius r  57.3

3 More About Radians  Comparing degrees and radians  Converting from degrees to radians

4 Angular Displacement  Axis of rotation is the center of the disk  Need a fixed reference line  During time t, the reference line moves through angle θ

5 Angular Displacement, cont.  The angular displacement is defined as the angle the object rotates through during some time interval   The unit of angular displacement is the radian  Each point on the object undergoes the same angular displacement

6 Average Angular Speed  The average angular speed, ω, of a rotating rigid object is the ratio of the angular displacement to the time interval

7 Angular Speed, cont.  The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero  Units of angular speed are radians/sec  rad/s  Speed will be positive if θ is increasing (counterclockwise)  Speed will be negative if θ is decreasing (clockwise)

8 Average Angular Acceleration  The average angular acceleration,, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change:

9 Angular Acceleration, cont  Units of angular acceleration are rad/s²  Positive angular accelerations are in the counterclockwise direction and negative accelerations are in the clockwise direction  When a rigid object rotates about a fixed axis, every portion of the object has the same angular speed and the same angular acceleration

10 Angular Acceleration, final  The sign of the acceleration does not have to be the same as the sign of the angular speed  The instantaneous angular acceleration is defined as the limit of the average acceleration as the time interval approaches zero

11 Angular Acceleration Angular acceleration α measures how rapidly the angular velocity is changing: Slide 7-17

12 Linear and Circular Motion Compared Slide 7-18

13 Linear and Circular Kinematics Compared Slide 7-19

14 Sign of the Angular Acceleration Slide 7-20

15 Relationship Between Angular and Linear Quantities  Displacements  Speeds  Accelerations  Every point on the rotating object has the same angular motion  Every point on the rotating object does not have the same linear motion

16 Centripetal Acceleration and Angular Velocity  The angular velocity and the linear velocity are related (v = ωr)  The centripetal acceleration can also be related to the angular velocity

17 Vector Nature of Angular Quantities  Angular displacement, velocity and acceleration are all vector quantities  Direction can be more completely defined by using the right hand rule  Grasp the axis of rotation with your right hand  Wrap your fingers in the direction of rotation  Your thumb points in the direction of ω

18 Velocity Directions, Example  In a, the disk rotates clockwise, the velocity is into the page  In b, the disk rotates counterclockwise, the velocity is out of the page Righty-tighty Lefty-loosey

19 Centripetal and Tangential Acceleration Slide 7-22

20 Force vs. Torque  Forces cause accelerations  Torques cause angular accelerations  Force and torque are related

21 Torque  The door is free to rotate about an axis through O  There are three factors that determine the effectiveness of the force in opening the door:  The magnitude of the force  The position of the application of the force  The angle at which the force is applied

22 Torque, cont  Torque, , is the tendency of a force to rotate an object about some axis   = Fr   is the torque  F is the force  symbol is the Greek tau  r is the length of the position vector  SI unit is N. m

23 Interpreting Torque Torque is due to the component of the force perpendicular to the radial line. Slide 7-25 Torque is a vector quantity The direction is perpendicular to the plane determined by the position vector and the force

24 A Second Interpretation of Torque Slide 7-26 Fsin 

25 Signs and Strengths of the Torque Slide 7-27 If the turning tendency of the force is counterclockwise, the torque will be positive If the turning tendency is clockwise, the torque will be negative

26 Multiple Torques  When two or more torques are acting on an object, the torques are added  As vectors  If the net torque is zero, the object’s rate of rotation doesn’t change

27 General Definition of Torque  The applied force is not always perpendicular to the position vector  The component of the force perpendicular to the object will cause it to rotate  When the force is parallel to the position vector, no rotation occurs  When the force is at some angle, the perpendicular component causes the rotation

28 General Definition of Torque, final  Taking the angle into account leads to a more general definition of torque:   Fr sin   F is the force  r is the position vector   is the angle between the force and the position vector

29 Lever Arm  The lever arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force  d = r sin 

30 Net Torque  The net torque is the sum of all the torques produced by all the forces  Remember to account for the direction of the tendency for rotation  Counterclockwise torques are positive  Clockwise torques are negative

31 Checking Understanding The four forces shown have the same strength. Which force would be most effective in opening the door? A. Force F 1 B. Force F 2 C. Force F 3 D. Force F 4 E. Either F 1 or F 3 Slide 7-23

32 Answer The four forces shown have the same strength. Which force would be most effective in opening the door? A. Force F 1 B. Force F 2 C. Force F 3 D. Force F 4 E. Either F 1 or F 3 Slide 7-24

33 Moment of Inertia  The moment of inertia, I, of a point mass is equal to mass of the object times the square of the distance from the object’s axis of rotation.  SI units are kg m 2  Applying Newton’s 2 nd Law results in  

34 Moment of Inertia  The moment of inertia of an object is the rotational equivalent to the mass of the object in a linear motion.  Ex. For linear motion, the heavier the mass the more difficult it is to get it to move. In rotational motion, a high I, moment of inertia, means that it is difficult to get the object to rotate on an axis.  The size of the moment of inertia, depends on the radius of rotation from the center axis and the distribution of mass around the axis of rotation.

35 Moment of Inertia  If the radius length is large, it will be more difficult to get the mass to rotate which indicates a higher moment of inertia. So if we apply a torque closer to the axis of rotation, it will be easier to cause the rotation to occur.  If the mass is distributed closer to the axis of rotation, the rotation will be easier to start and the Moment of Inertia will be less I 1 > I 2

36 Moments of Inertia for Various Objects ObjectLocation of AxisDiagramMoment of Inertia Equation Thin hoop of radius r Through central diameter Solid uniform cylinder of radius r Through center Uniform Sphere of radius r Through center Long uniform rod of length l Through center Long uniform rod of length l Through end Thin rectangular plane of length l and width w Through center

37 Newton’s Second Law for a Rotating Object  The angular acceleration is directly proportional to the net torque  The angular acceleration is inversely proportional to the moment of inertia of the object

38 More About Moment of Inertia  There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object.  The moment of inertia also depends upon the location of the axis of rotation

39 Moment of Inertia of a Uniform Ring  Image the hoop is divided into a number of small segments, m 1 …  These segments are equidistant from the axis

40 Other Moments of Inertia

41 Example, Newton’s Second Law for Rotation  Draw free body diagrams of each object  Only the cylinder is rotating, so apply  = I   The bucket is falling, but not rotating, so apply  F = m a  Remember that a =  r and solve the resulting equations

42 Torque and Equilibrium  First Condition of Equilibrium  The net external force must be zero  This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium  This is a statement of translational equilibrium

43 Torque and Equilibrium, cont  To ensure mechanical equilibrium, you need to ensure rotational equilibrium as well as translational  The Second Condition of Equilibrium states  The net external torque must be zero

44 Equilibrium Example  The woman, mass m, sits on the left end of the see- saw  The man, mass M, sits where the see-saw will be balanced  Apply the Second Condition of Equilibrium and solve for the unknown distance, x

45 Axis of Rotation  If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque  The location of the axis of rotation is completely arbitrary  Often the nature of the problem will suggest a convenient location for the axis  When solving a problem, you must specify an axis of rotation  Once you have chosen an axis, you must maintain that choice consistently throughout the problem

46 Notes About Equilibrium  A zero net torque does not mean the absence of rotational motion  An object that rotates at uniform angular velocity can be under the influence of a zero net torque  This is analogous to the translational situation where a zero net force does not mean the object is not in motion

47 Solving Equilibrium Problems  Draw a diagram of the system  Include coordinates and choose a rotation axis  Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object  For systems containing more than one object, draw a separate free body diagram for each object

48 Problem Solving, cont.  Apply the Second Condition of Equilibrium  This will yield a single equation, often with one unknown which can be solved immediately  Apply the First Condition of Equilibrium  This will give you two more equations  Solve the resulting simultaneous equations for all of the unknowns  Solving by substitution is generally easiest

49 Example of a Free Body Diagram (Forearm)  Isolate the object to be analyzed  Draw the free body diagram for that object  Include all the external forces acting on the object

50 Example of a Free Body Diagram (Beam)  The free body diagram includes the directions of the forces  The weights act through the centers of gravity of their objects

51 Example of a Free Body Diagram (Ladder)  The free body diagram shows the normal force and the force of static friction acting on the ladder at the ground  The last diagram shows the lever arms for the forces

52 Center of Gravity  The force of gravity acting on an object must be considered  In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at a single point

53 Calculating the Center of Gravity  The object is divided up into a large number of very small particles of weight (mg)  Each particle will have a set of coordinates indicating its location (x,y)

54 Calculating the Center of Gravity, cont.  We assume the object is free to rotate about its center  The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm  For example,    m 1 g x 1

55 Calculating the Center of Gravity, cont.  We wish to locate the point of application of the single force whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles.  This point is called the center of gravity of the object

56 Coordinates of the Center of Gravity  The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object

57 Center of Gravity of a Uniform Object  The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry.  Often, the center of gravity of such an object is the geometric center of the object.


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