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Physics 101: Lecture 16, Pg 1 Physics 101: Lecture 16 Rotational Kinematics l Today’s lecture will cover Textbook Chapter 8.

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Presentation on theme: "Physics 101: Lecture 16, Pg 1 Physics 101: Lecture 16 Rotational Kinematics l Today’s lecture will cover Textbook Chapter 8."— Presentation transcript:

1 Physics 101: Lecture 16, Pg 1 Physics 101: Lecture 16 Rotational Kinematics l Today’s lecture will cover Textbook Chapter 8

2 Physics 101: Lecture 16, Pg 2 Rotational Kinematics l The motion of a rigid body about a fixed axis is described by using the same concepts as for linear motion (see C&J Chapter 2): Displacement (  ), Velocity (  ), Acceleration (  ) Angular Displacement: Identify the axis of rotation and choose a line perpendicular to this axis. Observe the motion of a point on this line. How can one define the change of position of this point during rotation about an axis ? Answer: Change of angle the line makes with a reference line: 

3 Physics 101: Lecture 16, Pg 3 Angular Displacement l Follow point P on line perpendicular to the rotation axis: t=t 0 : t=t f :   =0 ff -f-f Angular displacement:  =  f -    =  f  = -  f SI unit: radian (rad) 1 rad = 360 degrees/(2  )  (in rad) = arc length / radius = s/r counter clockwise clockwise reference line P

4 Physics 101: Lecture 16, Pg 4 Angular Velocity and Acceleration l With the concept of displacement in place we can now define angular velocity and acceleration to describe the motion of a rigid body rotating about an axis: Average angular velocity = angular displacement/elapsed time  ave =  /  t = (  f -  0 )/(t-t 0 ) Average angular acceleration = change in velocity /elapsed time  ave =  /  t = (  f -  0 )/(t-t 0 )

5 Physics 101: Lecture 16, Pg 5 Rotational Kinematics (with comparison to 1-D linear kinematics) Angular Linear And for a point at a distance R from the rotation axis: s = R  v T =  R  a T =  R See text: chapter 8 See Table 8.1 v 2 = v a  x

6 Physics 101: Lecture 16, Pg 6 Angular and Tangential Variables A point on a line perpendicular to the rotation axis at distance R from the rotational axis moves with a tangential speed (t 0 =0s,  0 =0) v T = s/t =  R/t =  R (  in rad and  in rad/s) and an average tangential acceleration a T = (v T0 – v Tf )/t = R (  0 -  f )/t =  R How does this relate to the case of uniform circular motion we discussed before ? Uniform motion = constant tangential speed and a T = 0 The change of direction of v T, however, results into a centripetal acceleration: a c = v T 2 /R =  2 R (=constant) Nonuniform motion = increasing/decreasing tangential speed Magnitude of the total acceleration is then given by a = (a T 2 + a c 2 ) 1/2

7 Physics 101: Lecture 16, Pg 7 Conceptual Question You and a friend are playing on the merry-go-round. You stand at the outer edge of the merry-go-round and your friend stands halfway between the outer edge and the center. Assume the rotation rate of the merry-go-round is constant. Who has the greatest angular velocity? 1. You do 2. Your friend does 3. Same CORRECT Since the angular displacement is the same in both cases. you

8 Physics 101: Lecture 16, Pg 8 Conceptual Question Who has the greatest tangential velocity? 1. You do 2. Your friend does 3. Same CORRECT This is like the example of the "crack-the-whip." The person farthest from the pivot has the hardest job. The skater has to cover more distance than anyone else. To accomplish this, the skater must skate faster to keep the line straight. v T =  R (  is the same but R is larger) you

9 Physics 101: Lecture 16, Pg 9 Concept Question Who has the greatest centripetal acceleration? 1. You do 2. Your friend does 3. Same CORRECT Things toward the outer edge want to "fly off" more than things toward the middle. Force is greater on you because you want to fly off more. Centripetal acceleration is a c = R  2 and you have the largest radius. you


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