Physics 2101 Section 3 March 10rd : Ch. 10 Announcements: • Exam #2 grade posted • Next Quiz is March 12 • I will be at the March APS meeting the week of 15 1519th. Prof. Rich Kurtz will help me. me Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101--3/ http://www.phys.lsu.edu/classes/spring2010/phys2101 http://www.phys.lsu.edu/~jzhang/teaching.html

Rotational Work and Energy We can compare p linear variables with rotational variables

x v a Δt F m

θ ω α

s = rθ vT = rω aT = rα

Δt

τ I

The same can be done for work and energy: For translational systems

W =F⋅x 1 KE = mv 2 2

For rotational systems

W = τ ⋅θ 1 KE = Iω 2 2

α ‐ into board

Example #2 Massless cord wrapped around a pulley of radius r and  mass MW (frictionless surface/bearings) and I / / W r2.  W=1/2M What is angular acceleration, α, of pulley (disc)? 1) What are forces on m1? 

2) What are forces on m2?  2) What are forces on m ?

xˆ : T1 − m1 g sin θ = m1 a yˆ : N − m1 g cos θ = 0

yˆ : T2 − m 2 g = −m 2 a ⇒ T2 = m 2 (g − a )

⇒ T1 = m1 (a + g sin θ )

NOTE: T1 & T2 are NOT equal

3) What are torques about wheel?   τ net = rT1 sin ( 90 ) + rT2 sin ( −90

τ net = τ 1 + τ 2

)

= R (T1 − T2 ) (+ zˆ )

NOTE: angular acceleration vector is in negative‐z direction 

4) Solve for α ?   IW α =

here

at = α r

⇒

(

(

1 2

1 2

)

IW α = r ⎡⎣(T2 ) − (T1 ) ⎤⎦

M W r 2 α = r ⎡⎣( m2 ( g − a ) ) − ( m1 ( a + g sin θ ) ) ⎤⎦

)

M W R 2 α = rg ( m2 − m1 sin i θ ) − r (α r )( m2 + m1 )

α=

2 g ⎡ m2 − m1 sin θ ⎤ ⎢ ⎥ r ⎣ M w + 2 ( m1 + m2 ) ⎦

a) b)

If Mw = 0, same as  problem 5‐43   If θ =90º, same as  problem 11‐55 (HW # 9) 

Summary: Effects of rotation From before

With Rotation

m = M

a = 23 g 2g(m 2 − m1 sin θ ) a= M w + 2(m1 + m 2 )

v=

4gd (M − m ) M W + 2(m + M )

v=

4 3

gd d Done with both

Done via force/torque

Done via energy remember this for later

IIn each of these cases: “translation” was separate from “rotation” h f th “t l ti ” t f “ t ti ” Pure translation + Pure rotation  

SHW#6: Three masses M, 2M and 3M are shown in  the figure with two pulley of moment of inertial I  and radius r. (a) What is the linear Acceleration?

T1 − Mg = Ma IIa (T2 − T1 )r = Iα = r T3 − T2 = 22Ma Ma

Ia (T4 − T3 )r = r 3Mg − T4 = 3Ma

T2 T1

Solve for T4 , then T3 ,-----If I did it right Mg a= ⎡I ⎤ + 3 M ⎢⎣ r 2 ⎥⎦ Check if I=0

F = 3Mg − Mg = 6Ma g a= 3

T3 T4

SHW#6: Three masses M, 2M and 3M are shown in  the figure with two pulley of moment of inertial I  and radius r. The coefficient of friction is μk and the  table is L m long.

T2 T1

T1 − Mg = Ma

⎞ ⎛ I T2 = gM (3 − 2 μ k ) − a ⎜ 2 + 5M ⎟ ⎠ ⎝r

Ia a (T2 − T1 )r = Iα = r T3 − T2 − 2 μ k Mg = 2Ma

⎡I ⎤ gM ⎢ 2 (2 − μ k ) + M (4 − μ k )⎥ ⎣r ⎦ T2 = ⎡I ⎤ + 3M ⎢⎣ r 2 ⎥⎦ ⎞ ⎛ I T3 = −a ⎜ 2 + 3M ⎟ + 3Mg ⎠ ⎝r

Ia (T4 − T3 )r = r 3M − T4 = 3Ma 3Mg 3M

Need v = rω & a = rα

T3 = −Mg M (1 − μ k ) + 3Mg 3M = Mg M (2 + μ k )

Mg (1 − μ k ) a= ⎡I ⎤ ⎢⎣ r 2 + 3M ⎥⎦

T3 T4

SHW#6: Three masses M, 2M and 3M are shown in  the figure with two pulley of moment of inertial I  and radius r. The coefficient of friction is μk and the  table is L m long. Let us use Energy.

T2

T3 T4

T1

We can define zero so E mech ( start ) = 0 After block 3 has moved a distance d we have 6 Mv 2 2 I ω 2 + − (3M − M ) gd E mech = 0 − 2 μk Mgd = 2 2 3Mv 2 Iv 2 − μk Mgdd = + 2 − Mgdd (use ( v = rω ) 2 2r 2 Mgd (1 − μk ) 2 : v = Mgg (1 − μ k ) 2 I v = 2ad : a = 3M + 2 I r 3M + 2 r v f + vi 2

t=d: t=

I⎞ ⎛ 2d ⎜ 3M + 2 ⎟ ⎝ r ⎠ Mgd (1 − μ k )

Same as before Mg (1 − μ k ) a= ⎡I ⎤ + 3M ⎢⎣ r 2 ⎥⎦

SHW#6‐problem #4: A M kg woman is in a  rotor with a  r m radius, laying against the  inner surface. The friction between the  woman's sweater and the surface is μs. (a)  The rotor begins spinning with a constant  angular acceleration , and when the rotor  reaches enough angular speed, the door  drops but the woman stays sticking to the  wall. (a) What is the minimum angular  velocity that will allow the woman not to fall  down when the door drops? (b) What's the  maximum angular acceleration that will not  make the woman slide along the wall when  the rotor is speeding up to the angular  velocity calculated in (a)?

at = α r v2 ar = r

(a) Calculate Normal Force Mv 2 gives Frictional Force FN = Mar = r ( ) Calculate Tangential (b) g Force 2 M Mv Ff = μs > Mg Mα 2 r FT = Mat = < μ s Mrω 2 = μ s Mr (α t )

r

Chapter 11 Rolling Torque, Rolling, Torque and Angular Momentum In this chapter we will cover the following topics: - Rolling of circular objects and its relationship with friction - Redefinition R d fi iti off torque t as a vector t to t describe d ib rotational t ti l problems bl that th t are more complicated than the rotation of a rigid body about a fixed axis - Angular momentum of single particles and systems of particles - Newton’s second law for rotational motion -

Conservation of Angular Momentum ω

Describe their motion: θ (t) → ω (t) → α (t) Iω 2 I→ → Torque 2 ω

Angular l Momentum M t = Iω I iωi = I f ω f

Understanding rolling with wheels Wheel moving forward with constant speed vcom s = θR

displacement:   translation→ rotation

vcom

These relationships define “smooth define  smooth rolling motion rolling motion”

a com

ds d (θR) = = = ωR dt dt

dvcom d (ωR) = = = αR dt dt

Only if NO SLIDING [smooth rolling]

At point P (point of contact), wheel does not move

Understanding rolling with wheels II

All points on wheel  All points on wheel  g move with same ω. All points on move to the right with same  move with same ω.  All points on  outer rim move with same linear  linear velocity vcom as center of  wheel speed v = vcom.

Combination of  pure rotation rotation” and  and “pure pure  “pure translation”

v =ω ×r Note at point P:  at point T: 

vector sum of velocity = 0 vector sum of velocity = 2vcom

(point of stationary contact) (top moves twice as fast as com)

Kinetic energy of rolling

1 2

I comω + 2

1 2

Mv

2 com

= KE tot

Note: rotation about COM and translation of COM combine for total KE Note: rotation about COM and translation of COM combine for total KE Remember:   vcom = ωr 

Question

A ring and a solid disc disc, both with ith radi radius s r and mass m, m are released from rest at the top of a ramp. Which one gets to the bottom first?

1. Solid disc 2. Ring (hoop) 3. both reach bottom at  same time

Question #2

Two solid disks of equal mass, but different radii, are released from rest at the top of a ramp. Which one arrives at the bottom first?

1. The smaller radius disk. 2. The larger radius disk. 3. Both arrive at the same time.

The equation for the speed of the a disk at the bottom  of the ramp is       4 3

gll sin i θ

Notice, it does not depend on the radius or the mass  of the disk!! 

Rolling down a ramp : Energy considerations Object, with mass m and radius r,  roles  from top of incline plane to bottom.  What  is v, a, and Δt at bottom h

ΔE mech = 0 AT BOTTOM ΔKE tot = −ΔU

(KE

rot +trans,COM 1 2

)

final

[

L sin θ = h

− (0)init = − (0) final − (mgh )init

] Using 1‐D kinematics

2 mvCOM + 12 I comω 2 = mgL sin θ

v 2 = v02 + 2aL

⎛v⎞ 1 + 2 ICOM ⎜ ⎟ = mgL sin θ ⎝r ⎠

v2 g sin θ a= = I ⎞ 2L ⎛ ⎜1 + ⎟ 2 ⎝ mr ⎠ AND

2

1 2

2 mvCOM

2 COM

v

⎛ ICOM ⎞ ⎜ m + 2 ⎟ = 2mgL sin θ ⎝ r ⎠ vCOM =

2gL sin θ ⎛ I ⎞ ⎜1 + ⎟ ⎝ mr 2 ⎠

θ

2L ⎛ I ⎞ t= ⎜1 + ⎟ 2 g sin θ ⎝ mr ⎠ AT BOTTOM

Compare different objects Assuming same work done (same change in U),  objects with larger rotational inertial have larger KE bj t ith l t ti l i ti l h l KErot and during rolling, their KEtrans is smaller.

KE tot = KE trans + KE rot

vCOM =

⎛ I com ⎞ = KE trans ⎜1 + ⎟ ⎝ mr 2 ⎠

22gL L sin i θ ⎛ I com ⎞ 1 + ⎜ ⎟ ⎝ mr 2 ⎠

Roll a hoop, disk, and solid sphere down a ramp ‐ what wins? Roll a hoop, disk, and solid sphere down a ramp  what wins? Object Hoop Moment o of inertia large → small

Disk

Sphere

sliding block  (no friction)

Rotational              Fraction of Energy in Inertia, Icom Translation Rotation

1mr 2 1

0.5

0.5

1 2

mr 2

0.67                 0.33

2 5

mr 2

0.71                 0.29

0

1

0

slowest

Δt bottom =

fastest

2L ⎛ I ⎞ ⎜1 + ⎟ g sin θ ⎝ mr 2 ⎠

Problem #1 A solid cylinder starts from rest at the upper end of the track as  shown What is the angular speed of the cylinder about its shown.  What is the angular speed of the cylinder about its  center when it is at the top of the loop? Using conservation of mechanical energy:

0=

[( mv 1 2

0 = ΔE mech = ΔKE tot + ΔU grav 2 com

][

+ 12 I com ω 2 )final − (0)init + (mg(2R) final − (mg(h))init

]

mg(h − 2R) = 12 m (ωr ) + 12 (12 mr 2 )ω 2 2

Rearranging yields

2mg(h g( − 2R)) = ω 2 1 + 12 2 mr

[

4g(h − 2R) 3r 2

ω=

Trans + Rot ⇒

NOTE: Compare with before NOTE:  Compare with before

v = ωr =

4 3

g(h − 2R) =

]

4 3

gd y

⇐ Rolling

v= m = M

4 3

gd y

Problem #1 A solid cylinder of radius 10 cm and mass 12 kg starts from rest  and rolls without slipping a distance of 6 m down a house roof and rolls without slipping a distance of 6 m down a house roof  that is inclined at 30º. Where does it hit? Using conservation of mechanical energy: 0 = ΔEmech = ΔKEtot + ΔU grav 0 = ⎡⎢ ⎣

( mv 1 2

2 com

+ 12 I comω 2

)

final

− (0 )init ⎤⎥ + ⎡⎣(0 )final − (mgh )init ⎤⎦ ⎦

mgL sin θ = m (v ) + 1 2

2

1 2

(

1 2

Rearranging yields

⎛ v⎞ mr ⎜ ⎟ ⎝ r⎠ 2

)

2

2mgL sin θ = v 2 ⎡⎣1 + 1 2 ⎤⎦ m vbottom,COM =

4 3

gl sin θ

Then just use kinematics (vox, vvoy…)) Then just use kinematics (v