Physics 201, Review 3

Important Notes: v This review does not replace your own preparation efforts v  Exercises used in this review do not form a test problem pool. Please practice more with end of chapter problems.

About Exam 3 q  When and where

§  Monday Nov 20th, 5:30-7:00 pm §  Same location as for the previous ones.

q  Format §  Closed book §  One 8x11 formula sheet allowed, must be self prepared. (Absolutely no sample problems, examples, class lectures, HW

etc.) §  20 multiple choice questions. §  Bring a calculator (but no computer). Only basic calculation

functionality can be used. q  Special needs/ conflicts:

§  All requests for alternative test arrangements should have been settled by now. (except for medical emergency)

§  All alternative test sessions are in our lab room, only for approved requests.

Chapters Covered q  Chapter 9: Linear momentum and collision q  Chapter 10: Rotation about fixed axis

q  Chapter 11: Angular Momentum

q  Chapter 12: Static equilibrium and elasticity

note: Section 7.9: Energy diagram/equilibrium is also covered at conceptual level. (Also please note that the current homework assignment has a two

week due time. BUT please make an effort to work it out Before the midterm 3)

Your TAs will offer again a Super Friday tomorrow

Basic Concepts and Quantities q  Momentum, Angular Momentum

§  Linear Momentum, Impulse §  Angular Momentum (magnitude and direction) §  Impulse, Torque…

q  Collisions: Elastic, Inelastic, Perfectly Inelastic q  Center of Mass / Center of Gravity q  Rotational Motion (1-axis)

§  Angular displacement (Δθ) / Velocity(ω)/Acceleration (α). §  Moments of Inertia §  Rotational Kinetic Energy §  Rolling w/o slipping

q  Conservation Laws §  Energy, Momentum, Angular Momentum

q  Static Equilibrium q  Elasticity

§  Young’s, Shear, Bulk Modulus.

Basic Techniques q  Calculate translational and rotational kinetic energy. q  Calculate moment of inertia using Ι = Σmiri

2 for discrete masses q  Calculate using basic rotational kinematical equations q  Relate linear and rotational kinematic quantities q  Deal with applications such as:

§  Moving/rolling on slope, hill, curved track (energy approach) §  Rolling without slipping §  Static equilibrium. (seesaw, ladder on wall, etc..) §  Elastic or perfectly inelastic collisions, simple explosions §  Simple Pulley System with non-zero pulley mass. §  ω change on Merry-go-round , figure skating (Ifωf = Iiωi)

q  Vector product: Magnitude and direction. Cross products of unit vectors.

q  Use Right Hand Rules

Review: Linear Momentum and Momentum Conservation

q  Linear Momentum

q  Impulse-Momentum theorem

q  Momentum Conservation:

∑=+++=+++≡ jpvmvmvmpppp ...... 332211321

(impulse) ItFppp extif

≡Δ=−=Δ

0== extif Fifpp ,

Review: Momentum And Kinetic Energy

q  Recall: KE = ½ mv2 and p = mv q  That is:

For same momentum, the larger the mass, the smaller the KE

Kinetic Energy

KE =12mv2 =

12

(mv)2

m=p2

2m

Review: Collisions q  Collision: An event in which two particles come close and interact

with each other by force. §  Momentum is conserved in collision: Pf=Pi (Per Impulse approx.) §  Kinetic Energy of the system may or may not be conserved:

•  Elastic: KEf = KEi

•  Inelastic: KEf ≠KEi Two extreme cases: Elastic and Perfectly Inelastic.

v  Formulas for 1-D only, will not exam 2-D collision beyond conceptual.

€

v1 f =m1 −m2

m1 +m2

v1i +2m2

m1 +m2

v2i =m1 =m2

v2i

v2 f =2m1

m1 +m2

v1i +m2 −m1m1 +m2

v2i =m1 =m2

v1i21

221121 mm

vmvmvv ff +

+==

Be sure to understand and be able to calculate energy loss in a Perfectly Inelastic Collision

A Simple but not Trivial Quiz q  A very massive particle M, initially at speed v, is making an

elastic (1-D) collision with a very light particle m (at rest). What is the speed of the particle m after the collision?

v, 2v, very large (~M/m), very large (~ M2/m2)

Review Exercise: Two Body System q  Consider a system of two 1.0 kg balls as shown

§  Q3.a: what is total momentum? A: 2.0kgm/s, B: 4.0 kgm/s, C: zero , D: -4.0 m/s

§  pA = MAvA = 1.0x2.0=2.0 kgm/s , pB = MBvB = 1.0x(-2.0)=-2.0 kgm/s p=pA+pB = 2.0 + (-2.0) =0

§  Q3.b: what is total kinetic energy? A:0.0 , B: 4.0 J, C: 8.0 J, D: 16.0 J §  EA = ½ MAvA

2 = ½ 1.0x2.02=2.0 J , EB = ½ MBvB2 =2.0 J also

E=EA+EB = 2.0 + 2.0 =4.0 J

A B 2.0 m/s 2.0 m/s

Exercise 1: Momentum Change A ball with original momentum +4.0 kg⋅m/s hits a wall and bounces straight back without losing any kinetic energy. The change in momentum of the ball is: A.  zero B.  +4.0 kgm/s C.  -4.0 kgm/s D.  +8.0 kgm/s E.  -8.0 kgm/s

Solution: Δp= pf – pi = (-4.0) – (+4.0) =-8.0 kgm/s (why pf=-4.0 kgm/s?)

Review: Center of Mass For a multi-particle system: m1, m2, m3, ... at r1, r2, r3, ... one can define: Ø  Total mass: M = Σ mj = m1 + m2 + m3 + ... Ø  Center of Mass (CM) position:

Ø  CM Velocity and Acceleration

Ø  Now think of CM as a virtual particle, it has M, r, v, a

m1

m2

m3

r1

r2

r3

MrmrmrmrCM...332211 +++

≡

CM

rCM

vCM ≡drCMdt

=m1v1 +m2

v2 +m3v3 +...

MaCM ≡

dvCMdt

=m1a1 +m2

a2 +m3a3 +...

M

€

F ext = M a CM

vCM =Pj∑M

Review Exercise: Find Center of Mass q  Find the CM for these 3 object system. (all masses same)

3300

321

332211 Lm

LmmmmmmxmxmxmxCM =

⋅+⋅+⋅=

++

++≡

3300

321

332211 Lm

mmLmmmmymymymyCM =

⋅+⋅+⋅=

++

++≡

Exercise 2: Conceptual Q q  At t=0, the center-of-mass (CM) velocity of a multi-particle

system is zero. Which of the following statement is true: §  The system’s kinetic energy must be zero §  The system’s momentum must be zero §  The system’s angular momentum must be zero §  The total external force on the system must be zero §  None of above

€

v CM ≡Δ r CM

Δt=

m1 v 1 + m2

v 2 + m3 v 3 + ...

M=Σ p j

M

Exercise 3: CM and Collision q  Consider two particles in 1-D collision. Before collision, the left

particle, of mass m1=1kg, moves to the right at v1=2m/s, the right particle, of mass m2=2kg, moves to the left at v2=-3m/s.

Ignore all external forces (Impulse approximation) Ø  What is the system’s CM velocity before collision?

Ø  What is the system’s CM velocity after collision? (answer in 30 sec)

Ø  Answer: still -1.3 m/s (quiz: why?)

Ø  Suppose the collision is elastic (or perfectly inelastic), what are the velocities of the particles after collision?

(Exercise after class)

€

v CM =m1 v 1 + m2

v 2m1 + m2

=1× 2 + 2 × (−3)

1+ 2= −1.3m /s

Review: Moments Of Inertial

€

I = miri2∑ (= r2dm)∫

See demo for rolling of wheels with different I

Review Quiz Order the following objects, all having the same R and M, according their moments of inertia around there respective axis as shown.

( 1=largest)

1

1 2

3 3

4

Linear and Rotational Motion q  Observe the similarity in math between linear and rotational motion.

q  For a rigid object, its motion = translation of CM + rotation about CM §  Especially, total KE = KECM + KErot

Linear (1D) Rotational (1-axis) x θ

v ω

a α

m IKE=½ mv2 KE=½ Iω2

F=ma τ=Ια

momentum p=mv

angular momentum L= Iω =(rxp)

F=dp/dt τ=dL/dt

Reminder: Rotational Kinematics and Dynamics

q  Rotational Kinematics (constant acceleration)

q  τ ≡ Fsinφ r q  Rotational Dynamics:

  Στ = I α I : Moment of inertia = Σ miri2

q  Mechanical Equilibrium

ΣF = 0 and Στ = 0

Reminder: Angular Momentum And Angular Momentum conservation

q  Angular momentum L ≡ rxP (L=ΣLj if multiple particle) §  For a rigid object about a fixed axis: L= Iω

q  Angular Momentum and Torque:

dL/dt = Στ

q  Angular momentum conservation:

if Στ=0 Lf = Li , (Ifωf = Iiωi for 1-axis)

recall: P=mv

Review: Angular Momentum And Rotational Kinetic Energy

q  Recall: KErot = ½ I ω2 and L = I ω q  That is:

For same angular momentum, the larger the moment of inertia, the smaller the KErot

Rotational Kinetic Energy

KErot =12

Iω 2 =

12

(Iω)2

I=

L2

2I

Review Exercise : Right Hand Rule q  A uniform disc is rotating about the z axis as shown. What is the direction

of its angular momentum? A.  +x B.  +y C.  +z D.  None of above.

v  Use your right hand to verify it is actually –z.

q  Further questions: Ø  For the above disc, assuming R, ω, M are known, are you able to get its (linear) momentum and its kinetic energy?

Exercise 4: Climbing Up a Ladder q  An 80 kg man is one fourth of the way up a 10 m ladder that is

resting against a smooth, frictionless wall. If the ladder has a mass of 10 kg and it makes an angle of φ=60° with the ground, find the force of friction of the ground on the foot of the ladder

Ø  What is the friction from ground. A: 780N, B:200N, C:50N, D: 141N Solution: §  Identify external forces (NF,NW, m1g,m2g, fs), draw FBD §  Choose ground support as the reference point §  τNW = NWsinφ L §  τm1g =- m1gsin(90o-φ)L/4, τm2g =- m2gsin(90o-φ)L/2 §  τNF = τfs =0 §  Στ=0 NWsinφ L – (m1/2+m2)gcosφL/2=0 §  NW=(m1/2+m2)g/(2tanφ) = 141N §  also: 0= ΣFx= Ffriction-NW Ffriction= NW = 141 N to the right

NW

NF

Fs

m2g

m1g

Exercise 5: Hanging a Sign q  A sign of mass M is hung 1 m from the end of a 4 m long beam

(mass m) as shown in the diagram. The beam is hinged at the wall. What is the tension in the wire?

Solution: §  Identify external forces: (N,F, mg,Mg, T), draw FBD §  Choose the hinge as the reference point §  τmg = - 2mg , τMg =- 3Mg, τT = 4Tsin(30o) = 2T, τF = τN =0 §  Στ=0 -2mg -3Mg + 2T =0 T= (m+1.5M)g

SIGN

wire

θ = 30ο

1 m θ = 30ο

2 m Mg

mg

T N

F

3 m

4 m

Review Exercise: Jumping On Merry-Go-Round

q  A freely spinning Merry-Go-Round of mass mmgr =100 kg and radius Rmgr =2m has an initial angular speed ωi =6 rev/min. After a child of mass mc = 25kg jumps on it at the edge as shown, what is the new ω (Idisc= ½ mR2)?

Solution: free spinning = no torque Lf=Li Li = Imgrωi = ½ mmgrRmgr

2 ωi Lf = (Imgr + Ichild )ωf =(½ mmgrRmgr

2 + mcRmgr2 ) ωf

à  ωf = ½ mmgrRmgr

2 / (½ mmgrRmgr2 + mcRmgr

2 ) ωi = ½ mmgr/ (½ mmgr+ mc

) ωi = 4 rev/min

Exercise 6: Rolling w/o Slipping Down a Slope

q  A uniform wheel (or disc, or sphere) of mass M , radius R, and moment of inertia I=MR2 is rolling down a slope without slipping as shown. (θ=30o)

Calculate its CM acceleration. q  Solution: Ø  Step 1: FBD as shown Ø  Step 2: Set up axis as shown Ø  Step 3: Dynamics for CM (x direction): mgsinθ – fs = maCM Ø  Step 4: Dynamics for rotation: -fsR =- Iα Ø  Step 5: rolling w/o slipping: Rα=aCM Ø  Solve for unknowns: aCM=g/4, fs=mg/4

mg

N

θ

Same Exercise in lecture 19: Rolling w/o Slipping Down a Slope

q  A uniform disc (or wheel, or sphere) of mass M , radius R, and moment of inertia I is rolling down a slope without slipping as shown. Calculate its CM acceleration.

q  Solution: Ø  Step 1: FBD as shown Ø  Step 2: Set up axis as shown Ø  Step 3: Dynamics for CM (x direction): mgsinθ – fs = maCM Ø  Step 4: Dynamics for rotation: -fsR =- Iα Ø  Step 5: rolling w/o slipping: Rα=aCM Ø  Solve for unknowns:

mg

N

θ

aCM =gsinθ

1+ ImR2

, fs =mgsinθmR2

I+1

Exercise 7: Rolling w/o Slipping Down a Slope

q  Consider two spheres A and B of the same radius and mass rolling down the same slope without slipping. Sphere A is uniformly solid while sphere B is a uniform shell (i.e. hollow inside). Which one rolls faster? (Ignore air resistance)

A, B, Same

Exercise 8: Rotational and Linear Motion A solid cylinder (I = MR2/2) has a string wrapped around it many

times. When I release the cylinder, holding on to the string, the cylinder falls and spins without slipping as the string unwinds.

Ø  What is the downward acceleration of the cylinder as it falls? (g=9.8m/s2)

Solution:

Rotational:  τ=TR= Iα = ½ MR2α àT=  ½  Ma  (Note:  αR=a  is  used)  Translational:  Mg-­‐T  =Ma    Solve:  a  =  2/3  g  =  6.5  m/s2        

Review: Deformation and Elasticity q  Small deformation (strain under small stress):

Strain = Stress / (Elastic modulus) Ø  There are three general types of stress/strain:

tensile shear bulk

€

Y ≡tensile stresstensile strain

=F /AΔL /L

€

S ≡shear stressshear strain

=F /AΔx /h

B ≡ volume stressvolume strain

=ΔF / AΔV /V

= ΔPΔV /V

Good Luck !