Chapter 11 Rolling torque and angular momentum Smooth

Chapter 11 Rolling, torque, and angular momentum

Smooth rolling • Smooth rolling – object is rolling without slipping or bouncing on the surface • Center of mass is moving at speed vcom • Point P (point of momentary contact between two surfaces) is moving at speed vcom s = θR ds/dt = d(θR)/dt = R dθ/dt vcom = ds/dt = ωR

Rolling: translation and rotation combined • Rotation – all points on the wheel move with the same angular speed ω • Translation – all point on the wheel move with the same linear speed vcom

Rolling: pure rotation • Rolling can be viewed as a pure rotation around the axis P moving with the linear speed vcom • The speed of the top of the rolling wheel will be vtop = (ω)(2 R) = 2(ωR) = 2 vcom

Chapter 11 Problem 2

Friction and rolling • Smooth rolling is an idealized mathematical description of a complicated process • In a uniform smooth rolling, P is at rest, so there’s no tendency to slide and hence no friction force • In case of an accelerated smooth rolling acom = α R fs opposes tendency to slide

Rolling down a ramp Fnet, x = M acom, x fs – M g sin θ = M acom, x R fs = Icom α α = – acom, x / R fs = – Icom acom, x / R 2

Torque revisited • Using vector product, we can redefine torque (vector) as:

Angular momentum • Angular momentum of a particle of mass m and velocity with respect to the origin O is defined as • SI unit: kg*m 2/s

Newton’s Second Law in angular form

Angular momentum of a system of particles

Chapter 11 Problem 33

Angular momentum of a rigid body • A rigid body (a collection of elementary masses Δmi) rotates about a fixed axis with constant angular speed ω • Δmi is described by

Angular momentum of a rigid body

Conservation of angular momentum • From the Newton’s Second Law • If the net torque acting on a system is zero, then • If no net external torque acts on a system of particles, the total angular momentum of the system is conserved (constant) • This rule applies independently to all components

Conservation of angular momentum

Conservation of angular momentum

More corresponding relations for translational and rotational motion (Table 11 -1)

Chapter 11 Problem 51

Answers to the even-numbered problems Chapter 11: Problem 4 (a) 8. 0º; (b) more

Answers to the even-numbered problems Chapter 11: Problem 18 (a) (6. 0 N · m)ˆj + (8. 0 N · m) ˆk; (b) (− 22 N · m)ˆi

Answers to the even-numbered problems Chapter 11: Problem 26 (a) (6. 0 × 102 kg · m 2/s) ˆk; (b) (7. 2 × 102 kg · m 2/s)ˆk

Answers to the even-numbered problems Chapter 11: Problem 32 (a) 0; (b) (− 8. 0 N · m/s)tˆk; (c) − 2. 0/√t ˆk in newton·meters for t in seconds; (d) 8. 0 t− 3 ˆk in newton·meters for t in seconds

Answers to the even-numbered problems Chapter 11: Problem 42 (a) 750 rev/min; (b) 450 rev/min; (c) clockwise