Phys101 Test4 Review F20

Physics 101 Prof. Bob Ekey

Test #4 – Friday 11/20/20

Test Review

This is not all inclusive - hits salient points.

Content - Chapters 11, 12, 15 Need to know information from readings, lectures, HW + Labs. Problem solving and conceptual questions. Equation sheet provided – please provide feedback by noon Thurs. Test 1&2&3 equ sheet + front cover provided (no feedback possible)

Time to complete exam: 55 Minutes Note: Spending 15 minutes on a 3 pt Multiple Choice not recommended.

Part I. Multiple Choice 12 problems worth 3 points each = 36 possible points Partial credit may be given = Show your work

Part II. Short Answer 4 possible problems worth 12 points each Solve three of four problems given, your choice. All problems will have (at most) four parts (a) - (d) 3 problems x 12 points each = 36 possible points.

Partial credit may be given = Show work, equ, provide units, etc. Total available points: Part I : 36 Pts + Part II 36 Pts = 72 pts.

See website for special Combo T4

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Δ p = p f −

p i

Ch11: Impulse and Momentum Momentum, p – vector Units: kg m/s (Ns) Points in the same direction of velocity. Can have x & y components.

What’s so great about momentum? It’s conserved. Kinetic Energy can also be defined in terms of the momentum; using this, determine the kinetic energy of the truck.

Total Linear Momentum (Sort of like net force) useful in explosions/collisions

Total linear momentum for the car/truck system?

Change in momentum Magnitude, direction or both Difference between two momentum vectors.

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p = m ⋅ v

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P = p tot =

p ∑ = p 1 + p 2 + p 3 + ...

vector sum

Also useful in collisions

Momenta: plural of momentum

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K = 12mv

2 =(mv)2

2m=p2

2m

vector difference

Jx ≡ Fx (t)dtti

t f

∫ = Δpx

Impulse, J (vector) Units: Ns (kg m/s)

Impulsive force - force exerted during a small defined time interval Force can and will change with time. Has a set duration (small ∆t).

Multiply both sides by dt

Integrate both sides

Define impulse Jx as area under the Fx(t) curve between ti and tf.

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max = m dvxdt

= Fx (t)

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mdvx = Fx (t)dt

mdvxvi

v f

∫ =mvf −mvi = Fx (t)dtti

t f

∫

Compression Expansion

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Jx = FavgΔt

In terms of Average Force Favg

Newton’s II law reprise: F=ma gets a new look. Average Force Force at a given instant

An impulse delivered to a particle changes the particle’s momentum.

Note: Average force on the object, force not necessarily constant in time.

!Favg =

Δ!pΔt

=mΔ!vΔt

=m ⋅ !a!F(t) = d

!pdt

Jx = Favg,xΔt = Δpx

Yay derivatives?

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F net =

Δ P Δt

=d P

dt= ZERO

P f − P i = 0

P f =

P i

Conservation of Momentum

Conservation: The total momentum, P, of an isolated system is a constant. Interactions within the system do not change the system’s total moment.

By NIII all internal forces will add up to zero.

Collision (where you use mom con) Two or more objects interact (collide), may move independently/together pre/post collision

“One of the most important principles of physics”

Make your system big enough & include all things interacting.

Some flavors of collisions (explosions) Inelastic Collision: Objects move together post collision w/a common final velocity Explosions: Objects combined before “explosion” move off in different directions Elastic Collision: Objects bounce off one another (momentum + kinetic energy conservation)

Ptot = Constant, Individual particles momentum can change.

For collisions in 2D use Px & Py (components)

Assuming isolated system: Net external force system is zero.

Elastic collisions (Bouncing w/no damage)

Two balls m1 and m2 are traveling with velocity v1 and v2 and they collide elastically. Kinetic energy is conserved in elastic collisions.

To find an equation for their final velocities, we use both KE and momentum conservation equations.

Usually you’re given the masses and initial velocities of the objects. So, find the final velocities in terms of the given quantities.

Conservation of momentum

P : m1

v1o+ m2

v2o

before

= m1

v1 + m2

v2

after

Conservation of kinetic energy K : 1

2 m1v1o

2 + 12 m2v 2o

2 = 12 m1v1

2 + 12 m2v 2

2

v1 =

m1 − m2

m1 + m2

"

#$%

&'v1o

+2m2

m1 + m2

"

#$%

&'v2o

v2 =

2m1

m1 + m2

!

"#$

%&v1o

−m1 − m2

m1 + m2

!

"#$

%&v2o

Equations expanded from the text You will have these for the test

Chapter 11: Impulse & Momentum.

A 250 g ball collides with a wall. The figure shows the ball’s velocity and the force exerted on the ball by the wall. What is the vfx, the ball’s rebound velocity?

A 5000 kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s speed is 20 m/s What mass of water has collected in the car?

Ans: 5.0 x 102 kg

11.13

Setup the following problems

Ans: 6.0 m/s

11.14

One billiard ball is shot east at 2.0 m/s. A second identical billiard ball is shot west at 1.0 m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90º and sending it north at 1.41 m/s. What are the speed and direction of the first ball after the collision. Give the direction as an angle south east. A 20 g ball is fired horizontally with a speed v towards a 100 g ball hanging motionless from a 1.0 m long string. The ball undergoes a head-on perfectly elastic collision, after which the 100 g ball swings out to a max angle of 50º. What was v?

Ans: 1.7 m/s, 55º SE

11.48

11.63

Ans: 7.9 m/s

Two carts of identical mass are put back-to-back on a track. Cart A has a spring-loaded piston; Cart B is entirely passive. When the piston is released, it pushes against cart B, and.

(a) A is put in motion but B remains at rest. (b) Both carts are set into motion, with A gaining more speed than B. (c) Both Carts gain equal speed but in opposite direction (d) B is put in motion, but A remains at rest. A car accelerates from rest. It gains a certain magnitude of momentum and earth gains (a) a larger magnitude of momentum. (b) the same magnitude of momentum. (c) a smaller magnitude of momentum. (d) need more information.

Conceptual question

Answer: (c) Momentum must be conserved, and the momentum before is zero (they are both at rest), so the momentum afterwards must also be zero. So, the carts must travel off in opposite directions with the same speed.

Answer: (b). The change in momentum equals force times time. The force of the earth on the car is equal and opposite to the force of the car on the earth (NIII), and this occurs over the same time duration, so the momentum change of the earth and car are the same magnitude.

€

xcm =1M

mixii∑ =

m1x1 +m2x2 +m3x3 + ....m1 +m2 +m3

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ycm =1M

miyii∑ =

m1y1 +m2y2 +m3y3 + ....m1 +m2 +m3

Center of Mass - “balancing point”

Chapter 12: Rotation of a Rigid Body \“point at which all of the mass of an object or a system may be considered to be concentrated” “mass-weighted center of the object”

Describe the motion of a many particle system w.r.t. the C of M. Why? Because it is easier, and the same physics apply.

A “free” object will rotate around its center of mass. Model object/system as if it were concentrated at multiple points

The center of mass of an object may not necessarily lie - at the center of the object - at the location of the most massive particle in the object - within the object

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Isys = I1 + I2 + I3 + ...

Rotational Kinetic Energy: Units: Nm or J

What’s the sign of rotational KE?

Moment of Inertia, I Rot. equiv. of mass Units: kg m2

If you know both objects moment of inertia, then just add them up.

Parallel axis-theorem. Moment of inertia about an axis that is parallel to one through its center of mass a distance, d, from c. of m.

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Krot = 12 Iω

2

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I = mii∑ ri

2

I = ICM + Md 2

Check the units

Moment of inertia depends on the axis of rotation, and the distance of the masses from the axis.

Moment of inertia for combined/unknown system

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I = mr2

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I = miri2

i∑

Notice they all have the same form. # times mass times length squared Moment of inertia…

Point Particle

Multiple Point

Particles

You will have this for the final

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τ ≡ rF sinφ

CCW Torques are positve CW Torques are negative

Torque (t) – can cause rotations Units: Nm (not J) Rotational equivalent of force. “effectiveness of force in causing rot.”

Depends on the magnitude of the applied force F The distance, r, from the point of application to the pivot The angle f between the force and radius vector

Vector How does a seesaw work? What’s a lever good for anyway?

magnitude

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τ = dF = (r sinφ)F

Torques and moment arms Alternatively the distance (d=r sin f) from the pivot point to the line of action (the line along which the force acts)

d is called the moment/lever arm

Angle between d and F is 90º

Newton’s II law for rotations

“An object’s angular acceleration is equal to the torque exerted on it divided by its moment of inertia (rotational mass). The angular acceleration is in the same direction as the torque.”

“If the net torque on an object is not zero, the object will undergo an angular acceleration.”

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τ net =

τ i∑ = I α I, moment of inertia (rotational mass)

Static Equilibrium (balanced and/or stable) Net force and Net torque equal zero

translational and rotational equilibrium condition

Unbalanced forces produce translational acceleration Unbalanced torques produce rotational acceleration

Stable Equilibrium – Any small displacement results in restoring force or torque, which tends to return the object to its original equilibrium position.

Unstable Equilibrium – Any small displacement results in a torque, which tends to rotate the object farther from its equilibrium position.

As long as the center of mass (CoM) still lies inside and above the objects original base of support, it will be stable.

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F net =

F i = 0

τ net = τ i∑∑ = 0

CoM CoM

Translation and Rotation Motion (how we move) Rigid Body – object where distance between particles are fixed

Translational motion – Linear motion Pure Translation- particles have same instantaneous velocity Rotational Motion – Rotational motion about a fixed axis Pure Rotation- particles have same instantaneous angular velocity

Rolling – Combination of both (Curve Ball, Slice, Fade, “English”) Axis of

Rotation

€

Krolling = 12 Icmω

2 + 12 Mvcm

2

€

Krolling = 12 Icm

vcmr

"

# $

%

& '

2

+ 12 Mvcm

2 = 12 Icmω

2 + 12 M (rω)2

€

vcm = rω

€

s = rθ

€

vcm = rω

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acm = rα

Total Mechanical Energy

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Kf +Uf = Ki +Ui = ETot (CONSTANT )

Rolling without slipping (we want static friction)

Can relate Translation and Rotation motion

Arc Length Velocity Acceleration

Kinetic Energy of a rolling body A rolling body has both translational and rotational kinetic energy

Total KE = Rotational KE + Translational KE of CoM

Using the no slipping condition you can write this as

Conservation of total ME – for no slip rolling systems

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Emech = K +USame as b4, except we have rotational KE

For a ball/cylinder - Center of mass over point of contact

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L = mr2 vtr

"

# $

%

& ' = mrvt

τ net =

ΔLΔt

= ZERO

ΔL =Lf −

Li = 0

Lf =

Li

Lf∑ =

Li∑   recall   L = Iω

Angular Momentum, L Units: kg m2/s

Points in the same direction as w (RHR)

Can rewrite Newton’s II law for rotations in terms of L.

“The angular momentum of an isolated system (net torque = 0) is conserved. The total initial and final angular momentum are equal”

L = I

ω

Vector

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τ net = I α = I Δ

ω Δt

=Δ(I ω )Δt

=Δ L Δt

=d L

dt

For a point particle

Notice to change L either change the angular velocity or the moment of inertia

Recall ice-skater. Ang mom con

Keep track of total “L” before and after

A 4.0-m-long 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building. A 70 kg construction worker stands at the far end of the beam. What is the magnitude of the torque about the point where the beam is bolted into place? A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0 m long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s?

Setup the following problems

Ans: 0.28 Nm ccw

12.26

Ans: 12.5 kNm

12.22

A 200g , 40 cm diameter turntable rotates on a frictionless bearing at 60 rpm. A 20 g block sits at the center of the turntable. A compressed spring shoots the block radially outward along a frictionless groove in the surface of the turntable. What is the turntable’s angular velocity when the block reaches the outer edge?

Ans: 50 rpm

12.79

Two wheels with fixed hubs, each having a mass of 1.0 kg, start from rest, and have the forces shown below applied to them. Assume the hubs and spokes are massless, so that the rotational inertia is I=mR2. In order to impart identical angular accelerations, how large must F2 be? (a) 0.25 N (b) 0.50 N (c) 2.0 N (d) 4.0 N

Two wheels initially at rest roll the same distance without slipping down identical inclined planes starting from rest. Wheel B has twice the radius but the same mass as wheel A. All the mass is concentrated in their rims, so that the rotational inertias are I=mR2. Which has more translational kinetic energy when it gets to the bottom? (a) Wheel A (b) Wheel B (c) The kinetic energies are equal (d) need more information

Assume R2=2R1

Answer: (c) The radius doubles, which quadruples the moment of inertia. The torque is doubled by the increase in radius. As the angular acceleration is the torque over the moment of inertia the result is to halve the angular acceleration unless the force is also doubled.

Answer: (c) For each wheel, the gain in total kinetic energy (translational plus rotational) equals the loss in gravitational potential energy. Because the two wheels have identical mass and roll down the same distance they both lose the same amount of potential energy. Both wheels also have the same ratio of translational to rotational kinetic energy, so their translational kinetic energies are the same.

Conceptual/MC Questions

Time [s]

Am

plitu

de

One Oscillation Period - T

Dis

pla

ce

me

nt

[m]

€

X( t) = A sin(ωt)

€

ω = 2πf =2πT

€

T =1f

Chapter 15: Oscillations

Period, T - time it takes for one oscillation Units: s Frequency, f - # of oscillations in one second Units: Hz

Angular Frequency, Units: rad/sec

Displacement – displacement from equilibrium (±x) Units: m

Amplitude, A – max displacement from equilibrium Units: m

Simple Harmonic Motion (SHM)

Harmonic Motion – anything that repeats itself at regular intervals

Simple – special case for sinusoidal oscillations

Things that undergo SHM are SH Oscillators (SHO)

+x

€

Fnet = Fsmanet = −kx

€

Fnet = −Fs +Fa = 0Fa = kx

€

a(t) = −kmx(t)

Horizontal Mass/Spring system

Hooke’s Law:

If you apply a Force of 10 N to stretch the system 10 cm from equilibrium, what is the spring constant of the spring?

After you let go what is the net force on the mass? Does it accelerate?

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Fs = −kx if xo = 0

k- spring constant

“restoring force”

o

Equilibrium

Stretched from Equilibrium

Friction Free

Direction of force opposite of

displacement

Acceleration proportional to the negative of

displacement

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x(t) = Acosωt = xmax cosωtv(t) = d

dt x(t) = −ωAsinωt = −vmax sinωt

a(t) = ddt v(t) = −ω2Acosωt = −amax cosωt

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xmax = Avmax =ωA

amax =ω2A

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T = 4.0 s ω =2πT

=πsrad / s

Sketch the motion

Use previous page. Slopes?

What’s the period? w?

Equations of motion. For an initial displacement A in the +x direction.

Amplitudes for equations of motion

Note: + to the right, - to the left

Vel

ocity

[m/s]

A

ccel

. [m

/s2]

2 4 6 8

2 4 6 D

ispla

cem

ent [

m]

2 4 6 8

8

Time [s]

Direction of force opposite of displacement

(rad/s) (s) (1/s or Hz)

Does period depend on amplitude or gravity?

(rad/s) (s) (1/s or Hz)

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T = 2π Lg

€

f =12π

gL

€

ω =gL

€

θ(t) =θo cosωt

Mass/Spring System: w, f, and T relations

Angular Frequency, w Frequency, f Period, T

Pendulum: w, f, and T relations (small angles ~ 10º) Angular Frequency, w Frequency, f Period, T

ω =

km

f = 12π

km

T = 2π mk

Can define equations of motion for angular displacement

But we’ll only deal with mass/spring system equations of motion

Do you need the mass of the pendulum bob?

Equations valid for small angles q<10º

vmax =kmA =ωA

Total Energy (ch10 review?)

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E = K +U = 12 mv

2 + 12 kx

2

€

E = 12 mvmax

2 = 12 kA

2

€

v = ± km (A

2 − x 2)

€

A = x2 + mk v

2x = ± A2 − mk v

2 )

Constant for conservative system Friction Free

Total Energy in system

Note: Useful Algebra (you are welcome) At what point in the motion is KE max? is this vmax or amax? At what point in the motion is PE a max? is this vmax or amax?

KE

PE

+

=

CONSTANT

Ener

gy [J

]

Time [s]

Mass/Spring velocity Mass/spring position

Mass/spring amplitude

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Fnet = Fsp −Fg = 0 Fnet = kΔL −mg = 0

€

Fnet( )y = Fsp( )y + Fg( )y = k(ΔL − y)−mg

= kΔL −mg( )− ky = −ky

€

ΔL =mgk

Vertical Mass/Spring Static Equilibrium (∆L stretch of spring)

Let the block oscillate around this equilibrium position

Be careful, value given may be from unstretched length, Not equilibrium.

Equations same as before… which is nice.

The two graphs shown are for two different vertical mass-spring systems. (a)What is the frequency of system A? When is the first time at which the mass has a max speed while traveling in the upward direction? (b)What is the period of system B? What is the first time at which the energy is all potential? (c)If both systems have the same mass, what is the ratio of kA/kB? A 200 g block hangs from a spring with spring constant 10 N/m. At t=0 s the block is 20cm below the equilibrium position moving upward with a speed of 100 cm/s. What are the block’s (a)Oscillation frequency (b) Distance from equilibrium when the speed is 50 cm/s. (c) Amplitude of oscillation?

Setup the following problems

Ans: (a) 1.1 Hz (b) 0.73 m/s (c) 0.245 m

15.46

15.32

Ans: (a) f=0.25 Hz, 3.0 s (b) T=6.0s, 1.5s (c) 9/4

€

TmoonTearth

=2π L gmoon2π L gearth

=gearthgmoon

=gearth16 gearth

= 6€

6

€

1/ 6

You and your trusty vertical mass-spring system are on the moon. Which statement about the system is false? Recall that the acceleration due to gravity on the moon is 1/6 of the value found on earth. (a) The oscillation period on the moon is the same as when you are on earth. (b) At equilibrium, the net force on the mass is zero. (c) At equilibrium on the moon the spring is stretched further than when on the earth. (d) The period is the same regardless of the amplitude of oscillation. Answer: (c) At equilibrium, the net force on the mass would be zero and the spring force would equal the gravitational force. The force due to gravity on the mass would be 1/6 less than if you were on earth. So, the spring would be stretched 1/6 as much as if it were on the earth. You also took a pendulum with you to the moon, by what would the period of the oscillation on the moon change by in comparison to the period on earth? (a) 1/6 (b) (c) 6 (d)

Conceptual/MC Questions

Ans: (b). Set up the ratio of the period on the moon and on earth. All but the acceleration due to graving cancel. Plug in the gravity on the moon in terms of the gravity on earth… solve.