Section 2 Chapters 5-8

Chapter 5 Energy

Conservation of Energy is one of the great laws of physics. No contradictions have been found if all forms of energy are considered.

Types of energy are kinetic, potential, electromagnetic, and two types of nuclear. Chemical, solar, are electromagnetic. All forms of heat energy are kinetic.

Work and energy are the same thing. Work is defined as force in the direction of displacement time displacement

W=Fdcos(θ) Where Fcos(θ) is the component of force in the displacement

System Force Displacement Work kms newtions meters joules cms dynes centimeters ergs english pounds feet foot pounds

How much work do I do in walking up to room MS205 if I weight 170 lbs and carry “stuff” weighting 5 lbs. Assume the second floor is 10 ft above the ground floor.

W= -(170+50) 10 = -1750 foot pounds. – indicate that you do the work.

Kinetic Energy

v2=v02 +2ax multiply each side by m/2 => ½ m v2=½mv0

2 +max =>½ m v2=½mv02 +Fx

We define kinetic energy as KE=½ m v2

KE –KE0= FX=W. Doing work changes the kinetic energy of an object.

See Applied Physics 5.1. page 121.

See Example 51. Page 121

Problem: A car 2000 kg accelerates from 0 to 55 mph in a distance of 1000meters.

• How much energy did the gas provide? • What was the average force applied by the engine?

Answer: 50 mph = 22 m/sec KE –KE0= W => ½ m v2 -0= ½(2000)(22) 2 = 484,000 J

W=Fd => 484000/1000=484 N

If you can get the energy back a reaction is conservative. Friction and external forces are non-conservative forces.

Physics Energy Bank Account Deposit/Withdrawals Checking Account Savings Account CD Friction Kinetic Energy Potential Energy Spring Energy External forces

Potential Energy is gravitational energy and depends on position.

Work done to lift an object. W=Fd =mgy = PE

Drop a weight, slide one down an incline or do loops, the velocity is the same at the bottom.

Do Quick Quiz 5.2 on page 131

Example of non-conservative frictional force

Problem: A skier slides down a frictionless slope of angle 30 degrees and height of 20 meter. At the bottom she hits snow with a coefficient of kinetic friction of 0.1. What distance will she travel from the bottom of the slope?

Answer: The initial total energy is potential and is mgh. The final energy is zero. Work done by friction is of non-conservative is W=fd = µNd = µmgd. So all the energy goes into friction as heat.

mgh= µmgd => h== µd => d=hµ = 20/0.1=200m

Problem: A 2 kg mass sits on a frictionless table and is attached to 4 kg mass hanging over the edge of the table via string and pulley. The 4 kg mass is released from rest and falls 1 meter.

What is its velocity?

Answer Pick the zero height, y=0, to be at the end of the fall 1 meter below the start and say d is the height of the table above this point. Total energy to start KE =0, PE 4kg mass =4g1, PE of 2kg mass =2gd. The total energy must remain the same. At the end KE= ½ (4) v2 + ½ (2) v2, PE of 4 kg mass is 0 PE of 2 kg mass is 2gd.

4g+2gd= ½ (6) v2 + 2gd => 4(9.8) =3 v2 => v= 3.6 ms/sec

Spring Energy = ½ kx2 Where f= kx and k is the spring constant.

Problem A spring toy has spring of spring constant 20 n/m and is compressed 1cm. It mass is 1 gr. When the spring is released how high will the toy jump.

Answer: 1/2kx2 =mgh => h=1/2 (20)(.01) 2 /9.8(.001)=8.2 cm

Power is the rate of doing work P= W/t Units J/sec =Watt

English ft lb/sec 1hp =550 ft lb sec.

Problem: If a student studies 10 hour under a 100 watt light bulk and electricity cost $.10 per kikowatt hour how much does it cost the student?

Answer 100x10 = 1kw hour $0.1

Problem A 3200 lb car goes form 0 to 60 mph in 4 sec. What horse power engine I needed.

Answer 60 mph = 88 ft sec and mass of the car is 3200lb/32 = 100 slugs

W =½ m v2 = ½ 100(88) 2=38,720 ft lb

P =W/t = 38720/4 = 9680 ft lb/sec

9680/550 = 176 hp

Chapter Momentum and Collisions 7

The second great conservation law physics is conservation of momentum.

Momentum p is p =mv, where p and v are vectors so we have three laws one for each direction.

From F = ma = mΔv/Δt => F Δt= mΔv =Δp = Impulse

In words force over time is called impulse and causes a change in momentum

In space, if you miss wrench thrown to you and it hit your nose does it hurt.

Did you ever wonder why they don’t make cars with very springy bumper?

Take to case a car of mass m hits a brick wall and stop and one where the car simply bounces off the wall. Assume the time of the collision is the same in both.

Case 1

Assume a car moving to the right F Δt= Δmv = m(vf-vi) –m(0-v) => F= -mv/ Δt

Case 2

Assume a car moving to the right F Δt= Δmv = m(vf-vi) =m(-v-v) => F= -2mv/ Δt

In traffic accidents three things kill you

1. Force > 90 kN, 20,000 lb breaks bone in legs 2. Acceleration 150gs > 4 ms or 50 gs > 60 ms 50% fatality 3. Surface trauma 340 kN/m2 50% fatality

Assume 165 lb , 75 kg person and a car going 60 mph, 27 m/sec stops in .01 sec 10 ms and chest area of .5 m2

1. F = Δmv/ Δt = 75 (0 -27)/.01 = 203 kN Broken bone 2. a= Δv/ Δt= -2700 m/sec2 2700/9.8 = 280 gs brain dead 3. Surface trauma = 203/.5 = 406 kN/m2 dead?

What if various means we increased the time of the collosion to .1 sec 100 micro sec 1. F = Δmv/ Δt = 75 (0 -27)/.. = 20.3 kN OK 2. a= Δv/ Δt= -270 m/sec2 2700/9.8 = 28 gs OK 3. Surface trauma = 203/.5 = 40.6 kN/m2 OK?

How do you increase the time of a collision? Seat belts and air bag and collapsible front-ends.

Collision In all collisions momentum is conserved mvi =mvf

See problem 25 in problem section Conservation of momentum pi =0 = Pf

75v=8x12 v= 1.28 m/ sec d=vt =120x1.28= 153m Three types of collisions

• Elastic Collision energy and momentum are conserved (hard no elastic balls) • Inelastic Collision only momentum is conserved • Perfect Inelastic Collision momentum is conserved and bodies stick together.

Problem: Consider a perfect inelastic collision in with a 1,800 kg car going 15 m/sec collides head on with a 900 kg car going -15 m/sec. What is the final velocity of the two cars is the stick together?

1,800 kg 900 kg

Solution:

 

What force does a person of mass m in each car feel? Assume a collision time of t sec.

F =Δmv/t

For the 1,800 kg car Δmv= m(15-5) so force is 10m/t

For the 900 kg car Δmv= m(-5-15) so force is -20m/t twice the force. That is why it is safer to be in a bigger car.

I give these equations for an elastic collision of masses m1 and m2 with out proof. They are for a

collision where m2 is at rest to start. Eg v2i =0

Consider 3 Cases and find v1f and v2f

1. m1 = m2 v1f =0, v2f = v1i 2. m1 <<< m2 v1f =- v1i, v2f = 0 3. m1 >>> m2 v1f = v1i, v2f = 2v1i

Problem:

If an elevator falls form 110 floors how far can it fall before the break come on if you can withstand 10 gs

Answer: Call the distance fallen initially -y1, v0y1= 0, a=-g

 

After the breaks come on the distance fallen is y2 and v0y2=v1

 

Chapter 7 Rotation Motion and Gravity

Measure angular displacement with θ Δθ = θ2- θ1

If we measure θ in radians then arc length traveled in a circle S =rθ. Where r is the radius of the circle.

r S

θ

Note C=2πr=θr=S=> 2π rad in a circle of 360 degrees. This is a conversion factor.

Convert 30 degrees to radians 30 degrees rad/degree 2π/360 = π/60=.52

 

Problem: A car’s engine is idling at 1000 rpm what is its angular velocity?

Answer: 1000 rev/min 2π rad/rev sec/min 60/1 =105 rad/sec =ω

Problem: A girl is riding a bike with the wheels turning 3.4 rad/sec. She decides to pedal harder to speedup and accelerates with angular acceleration 2 rad/sec/sec for 3 sec. The radius of the bikes wheel is 0.5 m.

1. What is her angular velocity after 3 seconds? 2. How fast is the bike traveling? 3. How many rad has the wheel turned?

4. How far has she traveled?

Answer:

1. ω= ω0 + αt = 3.5 + 2*3 =9.5 rad/sec 2. v= ωr=9.5*.5 = 4.75 m/sec =10 mph 3. θ = ω0t +1/2 αt2 = 3.5*3 + ½ 2 *9 = 19.5 rad 4. S= θr = 19.5 8.5 =9.75 m

What do 78, 45, 33 1/3 recall? Records and these are rpms

CDs work different. The international standard is for disk to pass the reading laser at 1.3 m/s so the CD must change its angular velocity as the heads move in and out.

Problem Find the rpm of a CD when the read head is at 2cm and 5.6 cm

Answer: v= ωr

1. v=1.3 m/sce r= .02 m => ω =1.2/.02 =65 rad/sec rpm = 65 rad/sec 1 Rev/2π rad 60 sec/min =621 rpm

2. v=1.3 m/sce r= .056 m => ω =1.2/.02 =23 rad/sec rpm = 23 rad/sec 1 Rev/2π rad 60 sec/min =220 rpm

Centripetal Acceleration

A body moving in a circle must have an acceleration towards the center. This is because the direction of the velocity is changing.

 

Problem Of car on a curve.

Answer N

fk

mg

Y direction a=0 F=ma => N-mg=0 => N=mg

x direction -fk = ma = m(- v2/r)

Nµk =m v2/r

mgµk =m v2/r

µk = v2/rg

Problem Of car on a banked curve.

fk N

Θ

mg

Y direction F=ma NcosΘ =mg

X direction F=ma NsinΘ - fk = m v2/r

Assume no friction at all. Then NsinΘ = m v2/r

Divide and we have tanΘ= v2/rg.

This is important of ramp banking on highways and for rate of turn in airplanes.

Designs a rollercoaster See Example 7.9 page 206.

Gravity

Kepler’s Laws

1. Orbits are ellipses with sun at focus 2. Law of areas 3. Law of radius and period (time for one rev)

Kepler’s law of circular orbit. Find velocity for given orbit

. 

Kepler’s third law for circular orbits

T is the time for 1 rev which is a distance of 2πr. Using s=vt, =vT

 

Take T= 24 hours = 86400 sec, mass of earth and G to find height of a geosynchronous satellite.

Chapter 8 Rotational Equilibrium and Rotational Dynamics

Torque τ =F(perpendicular and moment arm r. F=rFsinθ

R

F(perp) θ F

See example 8.3 page 232

Center for gravity Y

m1 m2

x1 x2

Problem 1 m1= 2kg, m2= 2kg, x1 = -2 m, x2 =2 m x (cm) = 0

€

xcom =x1m1 + x2m2

m1 + m2

xcom =−2(2) + 2(2)2 + 2

= 0

Problem 2 2kg, m2= 2kg, x1 = -1m, x2 =2 m x (cm) = 0

€

xcom =x1m1 + x2m2

m1 + m2

xcom =−1(2) + 2(2)2 + 2

=14

= .25 

See Example 8.6 page 237

ΣF=ma => ΣrF=rma =Στ =mrrα =Iα where I=mr2 Is the moment or inertia.

Example 8.9 page 242

Using the Moment of Inertia table for solid shapes.

Kinetic energy of rotation KE = ½Iω2

Example 8.12 page 247

 

€

v⇒ω

a⇒α

m⇒ I

Example 8.15 page 252

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Σ F = m a Σ τ = I α

KE = 12 mv 2 KE = 1

2 Iω 2

p = m v L = I ω

Conervation of Momemtumm v i = m v f Ii

ω i = I f

ω f