Physics 106 Lecture 4 Work, Energy, Rolling, SJ 7th Ed.: Chap 10.8 to 9, Chap 11.1

• Work and rotational kinetic energy • Rolling • Kinetic energy of rolling • Examples of Second Law applied to rolling

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Completed: Rotational variables and angular kinematics (for constant α) Dynamical variables: rotational inertia and torque Newton’s Second Law for rotation and applications to systems of rigid bodies

Today:i Generalize work and energy conservation to include rotation ΔW = ΔKE ‰ Object tiive: use power and si mpli ci itty of conservat tiion l aws ‰ Principal changes: ‰ A torque acting through an angular displacement does work ‰ Kinetic energy includes rotational terms Rolling motion: rotation + mass center translation + a constraint

In the near future: Vector representation of torques: G G G G G ‰ use to treat complex geometry, τi = ri × Fi τnet = ∑ τi torques with multiple directions, etc. all i

A new conserved vector quantity called G G G G G G G angular momentum l i = ri × pi pi = mivi Ltot = ∑ l i ‰ based on linear momentum applied to rotation all i G G dL A re-statement of the Second Law: τ = net dt

1 Work/energy for linear motion… Energy Conservation - Work-Kinetic Energy Theorem ΔK ≡ K − K = ΔW ...work done by external forces (including potentials) f iG G G G Work : dW ≡ F D ds ...integrate... ΔW = F D ds ∫path 1 2 Kinetic energy : K = mv point mass only, no rotation 2

Energy Conservation - Mechanical Energy

ΔEmech ≡ ΔK + ΔU = ΔWnon-conservative only ΔU ≡ Potential energy difference; e.g., gravity ΔU = mgΔh E ≡ K + U Mechanical Energy mech G dW G ds G G power ≡ = F D = F D v dt dt For rotating rigid bodies…add… Kinetic energy associated with the rotation: K = Kcm + Krot ...so... ΔK = ΔKcm + ΔKrot 1 2 1 2 ...where... ΔK = K − K = Iω − Iω rot f,rot 0,rot ∑ 2 f ∑ 2 0 Torques can do rotational work even when zero work is done on the mass center:

dWrot = τdθ (torque x angular distance) ΔWrot = τdθ ∫limits Work-KE Theorem is the same as before but KE and Work include rotation

Work done by a pure rotational motion Apply force F to mass at point r, causing rotation-only about axis. Displacement is only along θ (transverse) direction G ƒ Find the work done by force F applied to the object at P as it rotates through an infinitesimal distance dsds= r =dθ rdθ G G dW ≡ F D ds as for translation = F cos(90 - φ)ds = F rsin(φ)dθ

ƒ Only the transverse component of the force (along the displacement) does work – the same component that contributes to the torque: τ = F rsin(φ)

ƒ The radial component of the force ∴ dW = τdθ does no work because it is perpendicular to the displacement

2 Work-Kinetic Energy Theorem for pure rotation

As object rotates from θi to θf work is done by the torque. Integrate dW = τdθ

θf θf ΔW = ∫ dW = ∫ τdθ (= τΔθ if τ does not depend on θ) θi θi Apply the Second Law τ = I α dω dW = Iα dθ = I dθ = I ωdω dt ωf ΔW = I∫ ω dω I is constant for rigid bodies ωi 1 2 1 2 ∴ ΔW = 2 I ωf − 2 I ωi = Kf,rot − Ki,rot = ΔKrot

Instantaneous Power: Divide both sides of dW = τdθ by dt dW dθ P ≡ = τ = τω dt dt

Example: The power output of a certain car is advertised to be 200 hp at 6000 rpm. What is the corresponding torque?

Convert the power to Watts: 1hp ≡ 746 Watts

⎛ 746 Watts ⎞ 5 P = 200 hp = 200 hp ⎜ ⎟ = 1.49 x 10 Watts ⎝ 1hp ⎠ Convert the angular velocity ω to rad/s: ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ω = 6000 rev/min = 6000 rev/min ⎜ ⎟ ⎜ ⎟ = 628 rad/s ⎝ 1 rev ⎠ ⎝ 60 s ⎠ Apply P = τω and solve for the torque:

P 1491.49 x 105 N. m/s τ = = = 237 N.m ω 628 rad/s

3 Example: An electric motor attached to a grindstone exerts a constant torque of τ = 10 N.m. The moment of inertia of the grindstone is I = 2.0 kg.m2. The system starts from rest. a) Find the kinetic energy after 8 seconds. b) Find the work done by the motor during this time. c) Find the average power delivered by the motor. a) τ = Iα α = τ /I = 10.0 / 2.0 = 5.0 rad/ s2

ωf = ω0 + αt = 5.0x8.0 = 40 rad / s Should equal work ΔW done by motor ΔK = K = 1 I ω2 = 1 x 2.0 x (40.0)2 = 1600 J. f 2 f 2 b) First, angle turned through: 1 2 1 2 Δθ = ω t + αt = x 5.0 x (8.0) = 160 rad 0 2 2 Definition of work, constant torque

θf ΔW = ∫ τdθ = τΔθ = 10. 0 x 160 = 1600 J. θ i ΔW 1600 J. c) Average power: P = = = 200 Watts. av Δt 8 s. but instantaneous = 0 W. at t = 0 power is not P ≡ τω { = 200 W. at t = 4 s. constant: = 400 W. at t = 8 s.

Energy Conservation: for translation + rotation about mass center (cm) ΔK ≡ ΔK + ΔK = ΔW ΔW includes both conservative and tot rot cm non-conservative forces, treated as OR external to system

ΔWnc includes only non-conservative ΔEmech ≡ ΔKtot + ΔU = ΔWnc forces, ΔU contains the conservative forces

Comparison: pure translation versus pure rotation Pure Translation (Fixed Direction) Pure Rotation (Fixed Axis)

Position coordinate x Angular position θ

Velocity v = dx/dy Angular velocity ω = dθ/dt Acceleration a = dv/dt Angular acceleration α = dω/dt

Mass m Rotational inertia I

Newton's second law Fnet = ma Newton's second law τnet = Iα Work dW = Fdx Work dW = τ dθ

2 2 Kinetic energy Kcm = (m/2)v Kinetic energy Krot = (I/2)ω Power (constant force) P = F.v Power (constant torque) P = τω

Work–KE theorem ΔW = ΔK Same, include Krot ΔW = ΔK

4 A wheel rolling without slipping on a table

• The green line above is the path of the mass center of a wheel. ω • The red curve shows the path (called a cycloid) swept out by a point on the rim of the wheel .

• When there is no slipping, there are simple vcm relationships between the translational (mass center) and rotational motion.

s = Rθ vcm = ωR acm = αR

Rolling without slipping

What does it mean for a wheel to roll rather than slide? • “No slipping” • contact point “P” is stationary, o/w it would be sliding • friction at “P” is static friction fs ≤ μsN

•distance covered s1 = arc swept out s2 as wheel rotates by Δθ • Mass center moves alongg(,) while wheel rotates around an axle (axis, cm) • Friction at the point of contact “matches” the rotation rate to the mass center speed, providing torque.

time t1 later time t2 Connection between constant ω mass center velocity and the angular velocity of the wheel P+ P • from the translational motion: s1 = vcom Δt = s2 • from the rotation: Δθ s = r Δθ = r ω Δt = s r 2 1 s2 • NO SLIPPING IMPLIES: P+ vcm = rω = vtang s1=rΔθ

5 Rolling = pure rotation around CM + pure translation of CM

Combination motions Æ actual rolling motion • the portion of the wheel at the bottom (at point P) is stationary

• the point at the top (at point T) is moving at speed 2vcom, (fastest) a) Pure rotation b) Pure translation c) Rolling motion

Someone moving with As seen from the “lab” with the CM of the wheel rotation turned off, all totalG velocityG for aG point sees tangential speed points on the wheel would vlab = vtang + vcom vtang = ωr = vcom move at speed vcom for no slipping at center : vlab = 0 + vcom at bottom : vlab = −vcom + vcom = 0 at top : vlab = vcom + vcom = 2vcom

Rolling as pure rotation about contact point P • Complementary views – a snapshot in time • Contact point “P” is constantly changing

vtang = 2ωPR vA = ωP 2Rcos(ϕ)

A v = ω R φ cm P R ωP φ

P vtang = 0

•Stationary observer sees rotation about “P” – the instantaneous center of rotation – with angular velocity wp

vcm = ωPR ωP = ωcm αp = αcm

Angular velocity and acceleration are the same about contact point “P” or about CM.

6 Kinetic energy for rolling, using contact point P

Show that: Ktot = Krot + Kcm The total kinetic energy of a rolling object is the sum of the translational energy of its mass center of mass plus the rotational kinetic energy about its center of mass

ƒ Stationary observer ƒ Pure rotation about “P” (snapshot) 1 v = ω r K = I ω2 cm P 2 p R ω ƒ Apply parallel axis theorem P 2 IP = Icm + MR ωR = vcm P rolling condition ∴ K = 1 I ω2 + 1 Mv2 2 cm 2 cm

KE of rotation KE of overall mass about mass center motion as a center axis particle with mass M

Example: Use energy conservation to find the speed of the bowling ball as it rolls w/o slipping to the bottom of the ramp ƒ Rotation accelerates if there is friction between the sphere and the ramp ‰ Friction force produces the net torque and angular acceleration. ‰ There is no mechanical energy change because the contact point is always at rest relative to the surface, so no work is done against friction

ƒ Apply conservation of mechanical energy

Emech = constant = Ki + Ui = Kf + Uf

‰ Let Uf = 0 at the bottom of the plane ƒ Suppose ball is a solid sphere ‰ Ki= 0 at the top (start from rest) 2 ‰ Ui = Mgh at the top 2 Icm = MR ‰ ω R = v 5 f f 1 ⎡ ⎤ ⎡ 2 5 ⎤ 2 1 2 1 2 1 I 2 ⎢ + ⎥Mvf = Mgh K = I ω + Mv = ⎢ cm +M⎥v 2 ⎣ 5 5 ⎦ f 2 cm f 2 f 2 ⎢ 2 ⎥ f ⎣ R ⎦ 1/ 2 ƒ Apply energy conservation ⎛ 10 ⎞ ∴ v = ⎜ gh⎟ ⎡ ⎤ f ⎝ 7 ⎠ 1 I 2 K = ⎢ cm +M⎥v = Mgh f 2 ⎢ 2 ⎥ f ⎣ R ⎦

7 Example: A wheel rolling without slipping • pulled by constant horizontal force F applied at the CM to accelerate wheel

• friction force fs makes it roll: applies torque to wheel, causes angular acceleration linear 2nd Law motion of cm: ∑Fx,i = macm = F − fs ∑Fy,i = 0 = N − mg cm rotational 2nd Law r axis through cm: ∑ τcm,i = − fsr = Icmαcm CCW = + F no slipping acm = −αcmr = − αPr (also ) P constraint: Icmacm fs ∴ fs = r 2 Icmacm Icmαcm substitute: macm = - mrαcm = F − fs = F − = F + r 2 r solve for Fr Same result by placing rotation axis at P: α = − = α I is I + mr2, -Fr is net torque α : cm 2 P P cm cm Icm + mr a is CW • F/m is the no-fitifriction accelera tion 2 ⎡ ⎤ Fr F 1 • If I =0 then a = F/m find acm: acm = = ⎢ ⎥ cm , cm 2 m 2 2 Icm + mr ⎣⎢1 + Icm /mr ⎦⎥ • For a hoop, Icm = mr , a=F/2m • For solid sphere, acm = 5F/7m I α I • f Æ 0 if I =0 f = − cm cm = F cm s cm find f : s 2 α ω s r I + mr • If F=0, = 0, is constant, fs=0 cm • f is to the left μ f s Minimum s for f ≤ μ mg ⇒ μ = s no slipping: s s s,min mg

Example: Thread unwinding from a spool • pull thread with constant horizontal force F applied at the top

• no slipping friction force fs applies torque, causes mass center acceleration

linear 2nd Law F cm motion: ∑Fx,i = macm = F − fs ∑Fy,i = 0 = N − mg cm nd r rotational 2 Law τ = - Fr − f r = I α CCW = + axis through cm: ∑ cm,i s cm cm

P no slipping a = −α r = − α r (also ) constraint: cm cm P

fs macm = - mrαcm = F − fs ⎛ Icm ⎞ ADD Æ 2 F = −αcm⎜mr + ⎟ SUBSTITUTE: Icmαcm ⎝ r ⎠ - = F + fs r Same result by placing rotation axis at P: Solve − 2Fr 2, αcm = = αP IP is Icm+ mr -2Fr is net torque for αCM: 2 Icm + mr α is CW

2 ⎡ ⎤ • Result is twice the previous one Fin d 2Fr 2F 1 acm = = ⎢ ⎥ • For a hoop, I = mr2, a=F/m I + mr 2 m ⎢1 + I /mr 2 ⎥ cm acm: cm ⎣ cm ⎦ • For solid sphere, acm = 10F/7m macm = - mrαcm = F − fs ⎛ I ⎞ Find fs: cm Icmαcm ADD Æ − 2fs = αcm⎜ − mr ⎟ = −F − fs ⎝ r ⎠ r • negative fs, acm to the right ⎡ 2 ⎤ 2 μ mr − Icm • to the right when mr >Icm Minimum s: f = −F 2 s ⎢ 2 ⎥ • For a hoop, Icm = mr , fs=0 fs use f ≤ μ mg ⇒ μ = ⎣⎢mr + Icm ⎦⎥ • For solid sphere, f = -3F/7 s s s,min mg s • For cylinder, fs = -F/3

8 Example: Rolling on a ramp w/o slipping • place rotation axis at mass center - ramp applies torque to the edge of rolling object

• friction force fs is up the ramp, o/w wheel slides. torque is CCW around axis

apply linear 2nd Law ∑Fx = Max,cm = fs - Mgsinθ to cm motion: ∑Fy = May,cm = N − Mgcosθ = 0 apply rotational 2nd ∑ τcm = Icmαcm = fsR α, Icm Law for axis at cm: • all other forces have zero moment arm Æ zero torque

• acm is in –x direction for CCW (+) αcm no slipping a = −α R = − α R (also ) constraint: x,cm cm P

Solve for mass center acceleration:

Fg= Mg Icmαcm Icmax,cm fs = fs = − R R2

− gsinθ Icm ax,cm = fs = M(ax,cm + gsinθ) − ax,cm(M + ) = Mgsinθ Icm R2 (1 + ) MR2

Same result using • General result for motion on ramp

point P – the τP,net + MgRsinθ • acm Æ g sinθ if we let Icm Æ zero instantaneous αcm = = • a is negative, so f is positive (up) I I + MR2 cm s rotation axis: p cm

Cylinders Racing to the Bottom

‰ 4.1. Two hoops of the same mass roll down a plane inclined at an angle θ with the horizontal. Each travels the same distance. The radius of hoop B is twice the radius of hoop A. There is no slipping or sliding. In what order do they reach the bottom?

A. A reac hes the b ott om fi rst b ecause it h as th e great er accel erati on. B. A reaches the bottom first because it has a smaller moment of inertia. C. B reaches the bottom first because it experiences a larger torque. D. B reaches the bottom first because it travels a larger distance in one rotation. E. They both reach the bottom at the same time, because each has the same linear acceleration.

‰ 4.2. How does the answer change if both rolling objects are now solid spheres?

‰ 4.3. How does the answer change if object A is a solid sphere and B is a hoop, both with the same radius and the same mass?

− gsinθ a = 2 I = 2 MR2 x,cm I Icm = MR cm 5 (1 + cm ) MR2

9 Example: A wheel driven w/o slipping by a CW torque applied through it’s axle. • Wheel under torque applies force G to the road

• fs is reaction force that acts on the wheel due to static friction • No work is done by either force as point “P” has zero displacement linear 2nd Law F = ma = f cm motion: ∑ x,i cm s rotational 2nd Law cm acm r axis through cm: ∑ τcm = Icmαcm = - τ0 + fsr CCW = + a no slipping α = α = − cm P constraint: cm P r Note: if f = 0 wheel slips (e. g. spins on ice) G fs s then acm = 0 and α = - τ0 / I 2 SUBSTITUTE: Icmαcm = macmr − τ0 = − mr αcm − τ0

Solve − τ0 Same torque around “cm” and “P” αcm = α α α 2 since cm = p for CM: Icm + mr Equals moment of inertia IP around “P”

FIND fs – friction force needed to let wheel accelerate under given torque: f is to the right, drives a τ0 1 s cm fs = macm = If I = 0, f = τ /r r 1 + I / mr 2 cm s 0 cm 2 If Icm = mr (hoop), fs = τ0/2r Minimum μ to keep f a s f ≤ μ mg ⇒ μ = s = cm from slipping: s s s,min mg g

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