Exam 2 Review

8.02W08D1

Announcements

Test Two Next Week Thursday Oct 27 7:30-9:30 Section Room Assignments on Announcements Page

Test Two Topics: Circular Motion, Energy, Momentum, and Collisions

Circular Motion: Vector Description

Position

Angular Speed

Velocity

Speed

Angular Acceleration

Acceleration

rr(t) =Rr̂(t)

ra(t) =ar (t) r̂ + aθ (t)θ̂ =−vω r̂(t) + Rα θ̂(t)

ω ≡dθ / dt

α ≡dω / dt =d2θ / dt2

a

r= vω =Rω 2 =v2 / R

v = r

v(t)

Modeling the Motion: Newton’s Second Law

• Define system, choose coordinate system.

• Draw free body force diagrams.

• Newton’s Second Law for each direction.

• Example: x-direction

• Example: Circular motion

totalˆ : .2

x 2

d xF m

dt=i

2totalˆ : .r

vF m

R=−r

Strategy: Applying Newton’s Second Law for

Circular Motion

• Always has a component of acceleration pointing radially inward

• May or may not have tangential component of acceleration

• Draw Free Body Diagram for all forces

• mv2/r is not a force but mass times acceleration and does not appear on force diagram

• Choose a sign convention for radial direction and check that signs for forces and acceleration are consistent

Concept Question: Tension and Circular Motion

A stone attached to a string is whirled in a vertical plane. Let T1, T2, T3, and T4 be the tensions at locations 1, 2, 3, and 4 required for the stone to have the same speed v0 at these four locations. Then

• T3 > T2 > T1 = T4

2. T1 = T2 = T3 = T4

3. T1 > T2 = T4 > T3

4. none of the above

Dot ProductA scalar quantity

Magnitude:

The dot product can be positive, zero, or negative

Two types of projections: the dot product is the parallel component of one vector with respect to the second vector times the magnitude of the second vector

cosθ⋅ =A B A Br rr r

(cos ) Aθ⋅ = =A B A B BP

r rr r r(cos ) Bθ⋅ = =A B A B A P

r r rr r

Review: Potential Energy Difference

c

B

c

A

U d WΔ ≡− ⋅ =−∫F rr r

Definition: Potential Energy Difference between the points A and B associated with a conservative force is the negative of the work done by the conservative force in moving the body along any path connecting the points A and B.

cFr

Review: Examples of Potential Energy with Choice of Zero

Point(1) Constant Gravity:

(2) Inverse Square Gravity

(3) Spring Force

U ( y) =mgy

U ( y =0) =0

U (r) = −

Gm1m2

r

U (r0=∞) =0

U (x) =(1 / 2)kx2

U (x =0) =0

Work-Energy Theorem: Conservative Forces

The work done by the total force in moving an object from A to B is equal to the change in kinetic energy

When the only forces acting on the object are conservative forces

then the change in potential energy is

Therefore

0

total total 2 20

1 1

2 2

fz

fzW d mv mv K≡ ⋅ = − ≡Δ∫ F r

r r

ΔU =−W =−ΔK

ΔU + ΔK =0

totalc=F F

r r

Change in Energy for Conservative and Non-conservative Forces

Force decomposition:

Work done is change in kinetic energy:

Mechanical energy change:

W =

rF ⋅d

rr

A

B

∫ = (rFc +

rFnc)

A

B

∫ ⋅drr =−ΔU +Wnc =ΔK

ΔK + ΔU = ΔEmech = Wnc

rF =

rFc +

rFnc

Strategy: Using Multiple IdeasEnergy principle: No non-conservative work

For circular motion, you will also need to Newton’s Second Law in the radial direction because no work is done in that direction hence the energy law does not completely reproduce the equations you would get from Newton’s Second Law

Constraint Condition:

Conservation

ΔK + ΔU = ΔEmech = Wnc = 0

r̂ : F

r= −m

vf2

R

0 at fN θ θ= =

Modeling the Motion Energy Concepts

Change in Mechanical Energy:

Identify non-conservative forces.

Calculate non-conservative work

Choose initial and final states and draw energy diagrams.

Choose zero point P for potential energy for each interaction in which potential energy difference is well-defined.

Identify initial and final mechanical energy

Apply Energy Law.

final

nc ncinitial

W d .= ⋅∫ F rr r

W

nc=ΔK + ΔU =ΔEmech

Bead on Track

A small bead of mass m is constrained to move along a frictionless track. At the top of the circular portion of the track of radius R, the bead is pushed with an unknown speed v0. The bead comes momentarily to rest after compressing a spring (spring constant k) a distance xf. What is the direction and magnitude of the normal force of the track on the bead at the point A, at a height R from the base of the track? Express your answer in terms of m, k, R, g, and xf.

Block Sliding off Hemisphere

A small point like object of mass m rests on top of a sphere of radius R. The object is released from the top of the sphere with a negligible speed and it slowly starts to slide. Find an expression for the angle θf with respect to the vertical at which the object just loses contact with the sphere. There is a non-uniform friction force with magnitude f=f0sinθ acting on the object.

Table Problem: Potential Energy Diagram

A body of mass m is moving along the x-axis. Its potential energy is given by the function U(x) = b(x2-a2) 2 where b = 2 J/m4 and a = 1 m .

a) On the graph directly underneath a graph of U vs. x, sketch the force F vs. x.

b) What is an analytic expression for F(x)?

Momentum and Impulse: Single Particle

• Momentum

SI units

• Change in momentum

• Impulse

• SI units

m=p vr r

mΔ = Δp vr r

f

i

t

t

dt≡∫I Fr r

[kg ⋅m⋅s-1] =[N⋅s]

[N ⋅s]

External Force and Momentum Change

The momentum of a system of N particles is defined as the sum of the individual momenta of the particles

Force changes the momentum of the system

Force equals external force, internal forces cancel in pairs

Newton’s Second and Third Laws for a system of particles: The external force is equal to the change in momentum of the system

rF =

rFi

i=1

i=N

∑ =drpi

dti=1

i=N

∑ ≡drpsys

dt

rF =

rFext

rF

ext=

drpsys

dt=

d(msys

rVcm)

dt=msys

rAcm

rp

sys≡

rpi

i=1

i=N

∑ =msys

rVcm

Strategy: Momentum of a System

1. Choose system

2. Identify initial and final states

3. Identify any external forces in order to determine whether any component of the momentum of the system is constant or not

i) If there is a non-zero total external force:

ii) If the total external force is zero then momentum is constant

rF

exttotal =

drpsys

dt

sys,0 sys,f=p pr r

Problem Solving Strategies: Momentum Flow Diagram

• Identify the objects that comprise the system

• Identify your choice if reference frame with an appropriate choice of positive directions and unit vectors

• Identify your initial and final states of the system

• Construct a momentum flow diagram as follow:

Draw two pictures; one for the initial state and the other for the final state. In each picture: choose symbols for the mass and velocity of each object in your system, for both the initial and final states. Draw an arrow representing the momentum. (Decide whether you are using components or magnitudes for your velocity symbols.)

Modeling: Instantaneous Interactions

• Decide whether or not an interaction is instantaneous.

• External impulse changes the momentum of the system.

• If the collision time is approximately zero,

then the change in momentum is approximately zero.

[ , ] ( )colt t

col ext ext ave col sys

t

t t t dt t+Δ

+Δ = = Δ =Δ∫I F F pr r r r

systemΔ ≅p 0rr

0coltΔ ;

Collision Theory: Energy

Types of Collisions

Elastic:

Inelastic:

Completely Inelastic: Only one body emerges.

Superelastic:

K

0sys =K f

sys

1

2m

1v

1,02 +

12

m2v2,02 +⋅⋅⋅=

12

m1v1, f2 +

12

m2v2, f2 +⋅⋅⋅

K

0sys > K f

sys

K

0sys < K f

sys

Elastic Collision: 1-DimConservation of Momentum and

Relative Velocity

Momentum

Relative Velocity

v

1,x ,i−v2,x,i =v2,x, f −v1,x, f

Table Problem: One Dimensional Elastic Collision:

Relative Velocity

Consider the elastic collision of two carts; cart 1 has mass m1 and moves with initial speed v0. Cart 2 has mass m2 = 4 m1 and is moving in the opposite direction with initial speed v0 . Immediately after the collision, cart 1 has final speed v1,f and cart 2 has final speed v2,f. Find the final velocities of the carts as a function of the initial speed v0 .