Ch10-1 Angular Position, Displacement, Velocity and Acceleration Rigid body: every point on the body moves through the same displacement and rotates through the same angle. Chapter 10: Rotational Kinematics and Energy  CCW +

CT1:A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular speed is A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine.

Angular Position Counterclockwise is positive

Radians  = s/r (a dimensionless ratio)

Angular Displacement

Average angular velocity  av = angular displacement / elapsed time  av = /t Instantaneous angular velocity  = lim /t t 0 Ch2-1 Angular Velocity Chapter 10: Rotational Kinematics and Energy  CCW +

P10.9 (p.308)

Average angular acceleration  av = angular velocity / elapsed time  av = /t Instantaneous angular acceleration  = lim   /t t 0 Ch2-1 Angular Acceleration Chapter 10: Rotational Kinematics and Energy  CCW +

Ch2-2 Rotational Kinematics Chapter 10: Rotational Kinematics and Energy  CCW +

CT2: Which equation is correct for the fifth equation? A.  =  0 +  B.  2 =   C.  2 =  0 +  D.  2 =  

Equations for Constant Acceleration Only 1.v = v 0 + at  =  0 +  t 2.v av = (v 0 + v) / 2  av = (  0 +  ) / 2 3.x = x 0 + (v 0 + v) t / 2  =  0 + (  0 +  ) t / 2 4.x = x 0 + v 0 t + at 2 /2  =  0 +  0 t +  t 2 /2 5.v 2 = v a(x – x 0 )  2 =   (  –  0 ) Assuming the initial conditions at t = 0 x = x 0 and  =  0 v = v 0 and  =  0 and a and  are constant.

1.  =  0 +  t 2.  av = (  0 +  ) / 2 3.  =  0 + (  0 +  ) t / 2 4.  =  0 +  0 t +  t 2 /2 5.  2 =   (  –  0 ) P10.20 (p.309) P10.22 (p.309)

Ch2-3 Connections Between Linear and Rotational Quantities s = r v t = r a t = r a cp = v 2 /r Chapter 10: Rotational Kinematics and Energy

CT3: A ladybug sits at the outer edge of a merry-go- round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s linear speed is A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine.

P10-29 (p.310)

CT4: P10.29c The force necessary for Jeff’s centripetal acceleration is exerted by A. gravity. B. Jeff. C. the vine. D. air resistance.

Ch2-4 Rolling Motion v = r if no slipping  = 0 if no friction Chapter 10: Rotational Kinematics and Energy

Rolling Without Slipping Constant v and  d = vt 2r = vt (2/t)r = v r = v recall that r = v t

P10.45 (p.311)

CT5: P10.45b If the radius of the tires had been smaller, the angular acceleration of the tires would be A. greater. B. smaller. C. the same.

Ch2-5 Rotational Kinetic Energy and Moment of Inertia For N particles: I = m i r i 2 and K = I 2 /2 Recall for translation K = mv 2 /2 Both translation and rotation: K = mv 2 /2 + I 2 /2 Chapter 10: Rotational Kinematics and Energy

Kinetic Energy of a Rotating Object of Arbitrary Shape: Rigid Body of N Particles

Table 10-1a Moments of Inertia for Uniform, Rigid Objects of Various Shapes and Total Mass M

Table 10-1b Moments of Inertia for Uniform, Rigid Objects of Various Shapes and Total Mass M

P10.52 (p.311)

CT6: P10.52b If the speed of the basketball is doubled to 2v, the fraction of rotational kinetic energy will A. double. B. halve. C. stay the same.

Ch2-6 Conservation of Energy W NC = E with K = mv 2 /2 + I 2 /2 Chapter 10: Rotational Kinematics and Energy

Problem Before After y=0 h vivi vfvf ii ff P10.60 (p.311)

CT7: P10.60b If the radius of the bowling ball were increased, the final linear speed would A. increase. B. decrease. C. stay the same.

CT8: In the race between the hoop and solid disk, which will arrive at the base of the incline first? A. hoop. B. disk. C. neither, it will be a tie.