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Phy 121 Section 9: Torque & Rotational Dynamics

f

Torque. (τ, Greek "tau") Torque is to rotation what force is to linear motion. Torque is found by

multiplying the force by the amount of leverage it has:

τ = l F

Problem 9 – 1: Find the torque in each position:

Weight of an object:

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Although actually distributed all over the object, its weight can be thought of as acting at its center

of gravity. (point where the object can be balanced.)

For a uniform object, C. G. is at its geometric center.

To find net torque: Add individual torques, + if counterclockwise, - if clockwise.

Problem 9-2: The lever weighs 100 N. The rope

is pulled with a 75.0 N force. What is the net

torque?

Ans: –11.2 N·m

Rotational counterpart of Newton's second law. Στ = Iα α in rad/s2

Example 9-a: A rope is wrapped around a disk of mass 1.50 kg and radius .200 m. If the rope is

under 10.0 N of tension and there is negligible friction, what is the disk's angular acceleration?

Ans: 66.7 rad/s2

Problem 9-3: 60.0 Nm is applied to this system. Find its angular

acceleration.

Ans: 2.00 rad/s2

Problem 9-4: Take a brake rotor to be a solid disk, with a mass of 4.00 kg and a radius of 13.0 cm.

Imagine that, with the car jacked up and the tire removed, this rotor is spinning at 500 rpm. The

brake is then used to stop it in .80 sec. How large is the friction torque from the brake?

Ans: 2.21 N·m

Problem 9-5: A 15.0 kg solid sphere of radius 7.70 cm is pivoted on a frictionless

axle through its center. A cord with a box on the end is wound around the

sphere's equator. The system is released from rest; after turning 63.0 radians

(about 10 revolutions), it reaches 50.0 rad/s. Find the tension in the cord.

Ans: 9.17 N

Translation Rotation examples:

x (or s) θ

v ω vf2 = vi

2 + 2aΔx

a α ωf2 = ωi

2 + 2αΔθ

F τ

m I KET = 1/2 mv2

KET KER KER = 1/2 Iω2

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Section 10: Statics (Objects at Rest):

1st condition of equilibrium: ΣF⃑ = 0⃑

(F = ma with a = 0) (or ΣFx = 0 and ΣFy = 0)

2nd condition of equilibrium: Στ = 0

(τ = Iα with α = 0)

Problem 10-1: (Same as 3-5) The tension in

the rope is the same from one end to the

other. The tightrope walker on the right

weighs 700 N. How large is the tension in

the rope? (There is an unknown friction

force which is approximately horizontal.)

Ans: 3.66 kN

Example10-a: How hard must you pull the rope to keep the 100 N

lever at rest?

Ans: 87.2 N

Problem 10-2: The see-saw is pivoted at its center,

and is perfectly balanced. Fred weighs 80 lb, Bob

weighs 95 lb, and Sue weighs 75 lb. What does Mary

weigh?

Ans: 127 lb

Problem 10-3: Fred and Bob are now sitting on this device, which

weighs very little. How hard does Mary have to push horizontally

at the point shown hold them up? (Mary is stronger than she

looks.)

Ans: 172 lb

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Problem 10- 4: The uniform metal plate is in a vertical

plane, and weighs 150 N. Rope A is pulled with 40 N.

How hard do you have to pull rope B to hold it in

equilibrium?

52.5 N

Example10 - b: The tension in the rope is 175 N. What

does the boom it is supporting weigh?

Ans: 163 N

Problem 10-5: The lever weighs 284 N. How hard do you have to

pull the rope to hold it in this position?

Ans: 81.4 N

Example10-c: An 800 N painter stands on the

uniform 200 N board. Find the tension in

ropes A and B.

Answer:

Free body diagram of board:

ΣFx = 0

no x forces

ΣFy = 0

TA + TB - 200 - 800 = 0

TA + TB = 1000

Στ = 0 will be true for any choice of pivot point.

About point B: - (TA)(4m) + (200N)(2m) + (800N)(1m) = 0

400 Nm + 800Nm = (TA)(4m)

1200 Nm = (TA)(4m)

1200 N∙m

4 m = TA

TA= 300 N

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ΣFy = 0 TA + TB = 1000

300 + TB = 1000

TB = 700 N

Problem 10-6: Two carpenters are carrying

40.0 lb ladder as shown. Find the force each

must lift with.

Ans: 12 lb, 28 lb

Example10-d: A 180 N ladder leans on the side of a house as shown. Find

H and V, the horizontal and vertical components of the force on the ladder’s

foot, and P, the horizontal force with which the wall pushes on the ladder’s

top.

Answer. Diagram:

ΣτA = 0 (Point A is bottom of ladder.)

(3.70 m)(P) – (.60 m)(180 N) + (0)(H) + (0)(V) = 0

3.70P – 108 = 0

3.70P = 108

P = 108/3.7 = 29.2 N

ΣFx = 0

H – P = 0

H = P so, H = 29.2 N

ΣFy = 0

V – 180 = 0 so, V = 180 N

Problem 10-7: This person's forearm

weighs 10.0 N. Find B, the force from the

biceps muscle, and H & V, the horizontal

and vertical components of the force from

the elbow.

Ans: 176 N, 95.9 N, -118 N

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Section 11: Fluids. The Gas Law.

Density: ρ = m

V m = mass, V = volume (ρ = Greek "rho")

Pressure: P = F

A F = force, A = area (PSI = lb/in

2, Pascal = N/m

2)

Absolute pressure = gauge pressure + atmospheric

total pressure pressure relative to atmospheric

(Atmospheric pressure is in formula sheet's table of units.)

Example11-a: A window is 117 cm by 144 cm, about 16 800 cm2. Find the force the atmosphere

exerts on it.

Ans: 170 kN

Hydrodynamics (moving fluids):

We will consider only smooth, steady flow without internal friction, for an incompressible fluid.

Comparing two points in a pipe,

Equation of Continuity: A1v1 = A2v2 or, Av = constant

Bernoulli's Equation: P1 + ½ρv12 + ρgh1 = P2 + ½ρv2

2 + ρgh2

or, P + ½ρv2 + ρgh = constant

Example 11-b: Find P in the constriction.

Ans: 6.40 x 104 Pa

(Where speed is higher pressure is lower. Examples: paint sprayer, carburetor, airplane wing.)

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Problem 11-1: The pipe has the same diameter at points A and B,

and is smaller at C. Point A is lower in elevation than B, which is

at the same level as C.

a. List the speeds vA, vB and vC from slowest to fastest.

b. List the pressures PA, PB and PC from lowest to highest.

Problem 11-2: Find the speed

and pressure of the water at

point 2.

Ans: 10.0 m/s, 410 kPa

Hydrostatics (fluids at rest):

Fill v = 0 into Bernoulli's equation. From P1 + 0 + ρgh1 = P2 + 0 + ρgh2 , if you increase P1, P2 goes

up the same amount:

Pascal's principle: The pressure increase at one point in an enclosed fluid equals the pressure

increase at every other point.

Problem 11-3: Input piston: radius = 1 cm; output piston: radius = 25cm. Input force = 80 N. Find the output force, assuming ideal conditions (No friction). Ans: 50 000 N Buoyancy: An object immersed in a fluid feels an upward force. (This is because pressure increases with depth: Pressure on bottom of object is more than pressure on top.) Archimedes' principle: Buoyant force = weight of displaced fluid = [(ρ of fluid)(V displaced)] g Example11-c: A 2.50 m

3 balloon plus the helium in it weighs 19.0 N. How hard do you have to pull

to hold it down? Ans: 12.6 N Problem 11-4: A 100 cm

3 rock, weighing 3.5 N on dry land, is weighed underwater. What does the

scale read? Ans: 2.52 N Problem 11-5: Compare buoyant force in previous problem to buoyant force on a 7.0 N object with the same volume.

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Temperature:

Water freezes: 0C, Water boils: 100C. Temperature corresponds to average kinetic energy per molecule. The colder a substance gets, the slower its molecules move. Absolute zero = the lowest temperature there is = -273C Kelvin scale: 0 K = absolute zero 273 K = 0C, 373 K = 100C, and so on.

The Ideal Gas Law:

V = volume

PV = nRT T = ABSOLUTE temperature

P = ABSOLUTE pressure (from collisions with gas molecules)

n = number of moles of gas (1 mole = 6.02 x 1023

molecules.)

R = 8.314 J/moleK

Example 11-d: You check your tire pressure at -7C, and the gauge reads 32 psi. If the tire is

heated to 38C, what will the gauge read then?

Ans: 39.9 psi

Problem 11-6: Short answer questions:

a. If a gas’s temperature is kept constant while the volume is reduced, what happens to P?

b. If V is kept constant, what happens to P if T is reduced?

Problem 11-7: Compression stroke in a diesel engine.

Before: The gas is at atmospheric pressure and 20°C.

After: The pressure is 66.3 atmospheres (absolute),

and the volume is 1/20th of what it was before. Find

the final temperature in C.

Ans: 698 °C

(The answer is hot enough to ignite the fuel. This is

why diesels don't need spark plugs.)

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Section 12: Heat

Thermal Expansion:

Linear expansion (change in length):

Coefficient of linear expansion: a property of the material. (Table in handout.)

Problem 12-1: An aluminum wire is exactly 40 m long when at 35C. Find its length at -20C.

Ans: 39.9472 m

Volume expansion: ΔV = Vo β ΔT where β 3α

Greek "beta": coefficient of volume expansion

HEAT = energy that flows from one object to another due to a difference in their temperatures.

Heat Units:

1 calorie = amount of heat needed to raise 1 gram of water by 1 C.

nutritionist's Calorie = 1 kcal = 1000 cal.

capital C

1 British Thermal Unit (BTU) = heat needed to raise 1 lb of water by 1F. (1 BTU = 252 cal.)

Mechanical equivalent of heat = conversion between Joules and calories. (1 cal = 4.186 J)

Example 12-a: A hacksaw does 100 J of work cutting a metal rod. How many calories is this?

Ans :

J

calJ

186.4

1)100( = 23.9 cal

Specific Heat Capacity (c) = amount of heat needed to raise 1 gram of the substance 1C.

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Q = m c ΔT Q = quantity of heat (cal or J), m = mass, T = temp (C)

Example 12-b: 100 cal is added to 50 g of steel at 20C. Find its final temperature.

Ans: 37.9C

By conservation of energy, in an insulated container ΣQ = 0

Example 12-c: 300 g of lead at 95C is placed in 100 g of water at 15C. What is their common

final temperature?

Ans: 21.7C

Problem 12-2: 90 grams of aluminum at 10C is placed in 200 grams of mercury at 120C. What is

their final equilibrium temperature?

Ans: 37.8C

Phase changes:

Adding heat does not change temperature while melting or boiling. (Heat energy goes into breaking

bonds between molecules, rather than increasing their speed.)

Heat of fusion (Lf) = heat per gram needed to melt the substance.

Q = mLf

Boiling: Q = mLv Heat of vaporization (Q = -mL if heat is lost.)

Example 12-d: Four grams of steam at 120C is cooled, becoming 50C water. How much heat

does it give off?

Ans: 2.39 kcal

Example 12-e: 4.0 g of steam at 120C is mixed with 70 g of 20C water. What is the mixture's final

temperature? (Assume all steam ends up as liquid.)

Ans: 54.0C

Problem 12-3: 50 g of ice at -30 is heated into 40C water. How much heat did it absorb?

Ans: 6700 cal

Problem 12-4: 50 g of ice at - 30C is placed in contact with 400 g of aluminum at 90C. What is

their final temperature? (Assume all ice ends up as liquid.)

Ans: 22.7C

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Review of sections 9 - 12:

1. At 20C, a steel ball bearing has a diameter of 2.000 cm, and sits on a 1.995 cm hole in a brass

plate. This system is then placed in an oven, making both metals expand. Above what temperature

will the ball be able to go through the hole?

Ans: 334C

2. Without any drawers, the filing cabinet weighs 220 N. Each

drawer and its contents weighs 95 N. "CG" stands for center of

gravity. With the bottom drawers out, as shown, how far out can you

pull the top drawer's center of gravity before the cabinet tips over?

Ans: 28.2 cm

3. Water enters a building through a three inch diameter pipe at 40.0

lb/in2, moving at .150 m/s. All of this water flows on through a half

inch diameter pipe in the basement, where the pressure is also 40.0

lb/in2. How much lower in elevation is the half inch pipe compared to the three inch pipe?

Ans: 1.49 m

4. A 7.5 Nm torque acts on a 3.0 kg solid sphere with a 35 cm radius for seven seconds. If it starts

from rest, how many radians does it turn?

Ans: 1250 rad 5. Short answer, 5 points each: a. A pair of sneakers is hung from the center of a clothesline to dry. Is the tension in the line

more, less, or the same as the total weight of the sneakers?

b.

i. List all of the objects shown which have no net force on them.

ii. List all of the objects shown which have no net torque about point P. c. The temperature of an ideal gas is reduced while the volume of its container stays constant.

What would the pressure be if it reached absolute zero? (Unlike a real gas, this one does not

liquefy before it gets there.)

d. The heat needed to raise one pound of water by 1F is called a _____________.

e. A meter stick hangs from a string tied to the 30 cm mark. A

weight hung from the end keeps it balanced, as shown. Which

weighs more: W, the meter stick, or are they the same?

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Section 13: Thermodynamics

First law of thermodynamics (a version of conservation of energy):

Q = ΔEint + W

ΔEint = increase in internal energy, Q = heat added, W = work done by system

Internal Energy, Eint = the total energy of the system's molecules. With a gas (but not always for

liquids and solids), Eint is proportional to the temperature:

ΔEint = (some constant)ΔT

Work, W:

In general, W = area under curve on P-V diagram

Problem 13-1: 200 J of heat is added to a gas while 150 J of work is done by pushing the piston

inward. How much does the internal energy change?

Problem 13-2: Find the work done:

Ans: -11 000 J

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Heat, Q: With a gas, it's more convenient to do specific heat on a per mole basis, rather than per

gram:

Q = n C ΔT C = molar heat capacity, or "molar specific heat"

C is measured in cal or J _

moleC moleC

CV = C at constant volume.

CP = C at constant pressure.

(At constant volume, all heat added raises the temperature. At constant pressure, only some heat

raises T; the rest leaves again as the expanding gas does work, so you need more heat per degree.)

Prob. 13-3: How much heat is needed to raise 2 moles of helium by 100C at (a) constant volume,

(b) constant pressure?

Ans: 2500 J, 4160 J

Thermodynamic Processes:

An isobaric process means constant pressure.

An isothermal process means ΔT = 0. (And therefore ΔEint = 0.)

graph: From PV = nRT, PV = constant

An adiabatic process means no heat flow. (Q = 0)

graph: PVγ = constant where γ = CP/CV

Example 13-a: Compression stroke in a diesel engine: Air at atmospheric pressure and is

compressed adiabatically to 1/20 of its original volume. Find the final pressure.

Ans: 66.3 atm

Earlier, we saw a temperature increase goes along with this. The same thing happens with

refrigerators, and other heat pumps: A substance is compressed, raising its temperature. It then

circulates through tubing on the outside of the refrigerator, radiating heat into the room. Then, it

expands, lowering its temperature, and circulates through tubing inside the refrigerator, absorbing

heat. It then returns to the compressor. Thus, heat is pumped from the inside to the outside. (This is

also aided by the refrigerant's changes of state.)

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Problem 13-4: Helium at 1000 kPa expands adiabatically to 18 times its original volume. Find the

final pressure.

Ans: 8.01 kPa

Second law of thermodynamics: Entropy doesn't spontaneously decrease. (ΔS 0)

Entropy, S: measures the randomness or "disorder" of a system.

("Disorder" means a more likely state; "order" means a less likely state. For example, for 100

pennies in a box, 50 heads and 50 tails has more entropy than 100 heads. This is because 50/50

corresponds to many more ways of arranging the pennies, and each way of arranging them is

equally likely to come up if the box is shaken.)

So, if you start with all heads, shaking them will often give 50/50, but if you start with 50/50,

shaking won't give all heads. Entropy won't decrease.

Second law, restated: Heat doesn't spontaneously flow from low temp to high temp.

(If a cold pan is placed on a hot stove, the thermal energy is unevenly distributed, just like an

uneven number of heads and tails. As the heat flows, making things even out, the system is

going into a more likely state.)

Heat Engines. (Devices which convert heat into mechanical work.)

They all take in heat, do work, and then give off heat:

Net work done per cycle = area enclosed on PV diagram.

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Prob. 13-5: Find the net work per cycle done by this

engine.

Ans: 4000 J

Efficiency = work out e = Wout

energy in Qh

Example: If fuel containing 1000 J is used to put out 300 J of work, what is the efficiency?

Ans: (300 J)/(1000 J) = .30 (or 30%)

Conservation of energy (1st law) says e < 100% for anything.

Devices which don't involve heat (levers, electrical transformers, etc.) have percent efficiencies that

sometimes make it well into the 90's. (Non-thermal forms of energy can, in principle, be converted

into each other with 100% efficiency.) But, heat is special because it is associated with

disorganized molecular motion. To convert heat into 100% mechanical energy would mean a

spontaneous increase in organization - like shaking pennies, and getting all heads.

So, the theoretical limit on a heat engine’s efficiency is lower than other devices because of the

second law:

For a heat engine: emax = 1 − Tc

Th (T = absolute temp.)

(Efficiency of the Carnot Cycle.)

(Friction and other shortcomings of the design cause further reductions in efficiency below this

theoretical maximum.)

Prob. 13-6: A steam engine's boiler has a temperature of 250C and the exhaust leaves at 105C.

The engine produces 150 J of work for every 1000 J in its fuel. Find:

a. the engine's efficiency.

b. the highest efficiency theoretically possible for it.

Ans: 15%, 27.7%

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Section 14: Special Relativity

Based on the Michelson - Morley experiment (see text) and also theoretical considerations, c (speed

of light in a vacuum) is the same to all observers.

Speed = (distance)/(time). So, for c to be the same, distance and time must be different in different

reference frames.

More time and distance between events in the frame where light has to catch up to the detector.

Time Dilation: Δt = γ Δt0 where γ = 1

√1− ( 𝑣

𝑐 )

2

Lorentz - Fitzgerald Contraction: L0 = γL

To keep straight which reference frame is which: In the moving frame, things should be shorter

(just along the direction of motion), and time should be slower.

To 3 digit accuracy, γ differs from 1 only above roughly 10% of c.

Example 14-a: A spaceship with a rest-length of 10 m is moving away from you at .99c.

a. What do you measure for its length?

b. How long, by your watch, does it take a clock on the ship to advance one second?

Ans: 1.41 m, 7.09 s

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Problem 14-1: The average lifetime of a stationary muon is 2.2 μs. If one is created 8.0 km above

sea level (by a cosmic ray hitting an air molecule), and it travels at .998c, find

a. its lifetime as seen by us, and

b. the distance to sea level in its reference frame.

Ans: 34.8 μs, 506 m

(The answers explain why they are detected at sea level, even though they shouldn't "live" long

enough to get there.)

Momentum: It’s actually γmv that’s conserved in collisions.

(While gravity depends only on m, how hard something hits you acts like the mass has

increased from m to γm.)

Energy:

Calculating the work needed to get from rest to a speed V gives:

KE = γmc2 - mc

2 (KE = 1/2 mv

2 is true only if v << c.)

KE = (γm - m)c2

So, if you set an object in motion, the energy gained equals the mass gained times a constant. This

is just like changing units: Inches times a constant equals feet; kilograms times a constant equals

joules. A kilogram, then, is just another unit of energy; mass and energy are equivalent: E = mc2

(Applies to all forms of mass and energy.)

Eo = mc2 = Rest energy (Energy equivalent of the rest mass.)

Etot = γmc2 = Total energy

KE = Etot - Eo .

In atomic physics, energy is often measured in electron-volts rather than joules. 1 eV = 1.6 x 10-19

J.

1 MeV = 106 eV.

Prob. 14-2: For an electron moving at .800c, what is its

a. rest energy in eV?

b. total energy in eV?

c. kinetic energy?

Ans: 5.11 x 105eV, 8.52 x 10

5eV, 3.41 x 10

5eV

Why no object can reach c: As v approaches c, γ m = m

√1− (v

c )

2 approaches m/0 = . Giving a

body kinetic energy is the same as giving it mass (in the sense of how it moves). As it nears the

- 47 -

speed of light, it gets too much inertia for further significant acceleration. Adding more energy

makes the particle "heavier" not faster.

Binding energy = the energy needed to pull an atomic nucleus apart.

When apart, the protons and neutrons add up to more mass than when stuck together, because the

potential energy added in pulling them apart is equivalent to mass.

Binding energy = (mass apart - mass together)c2

Example 14-b: A pion at rest (mπ = 2.4881 x 10-28

kg) decays into a muon (mμ = 1.8835 x 10-28

kg)

and an antineutrino (mν̅ ≈ 0.) The reaction is written π– → μ

– + �̅�. It can be shown from

conservation of momentum that 12.15% of the energy produced by this decay goes to the kinetic

energy of the muon and the rest goes to the antineutrino. Find (a) the kinetic energy of the muon.

(b) the speed of the muon.

Ans: 6.60 x 10-13

J, .271c

Ex. 14-3: In atomic mass units (1 u = 1.66 x 10-27

kg), the mass of a proton = 1.007825, the mass of

a neutron = 1.008665, and the mass of a 21H nucleus = 2.014102. What is the binding energy of the

21H nucleus? Give the answer in both joules and eV.

Ans: 3.56 x 10-13

J, 2.22 x 106 eV

Problem 14-4: What is the speed of a meter stick which is only 80 cm long?

.600 c

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Review for Final Exam

The test has eight parts each worth 25 points. The best seven you do will be counted. (Perfect score

= 175.)

The questions below were picked for being things people often need to work on more, not for

similarity to the actual test. You should review the entire course, not just the topics on this sheet.

1. A 120 lb boy is leaning back in his chair, as

shown. The chair weighs 30 lb. Find the force

on his foot from the wall.

Ans: 23.3 lb

2. A 1200 kg car drove up a hill, as shown. It

took 10 seconds to go from the initial position to

the final position. Ignore friction. Find

a. the work done by the engine.

b. the energy put out by the engine.

c. the engine’s average power output.

d. the average force due to the engine.

Ans: a) 52 800 J, c) 5280 W, d) 211 N

3. The steel block has a mass of 75 kg, and sits on a steel ramp.

What is the minimum force the rope must pull with to keep the block

from sliding away?

Ans: 39.9 N

4. A wheel slows down uniformly with an angular acceleration of -.650 rad/s2 for 10.0 seconds. If

it completes 8.50 revolutions during this 10 seconds, how fast was it spinning at the start?

Ans: 8.59 rad/s

5. A racquetball leaves the racquet going 20.0 m/s, 29° above the horizontal. How far above the

level of the racquet does it hit the wall, 8.00 m away?

Ans: 3.41 m

6. 200 grams of molten lead at its freezing point, 327C, is poured into 300 g of water at 23.5C, in

a sealed, insulated container. When the lead and water reach thermal equilibrium, they are at

33.5C. What is the heat of fusion of lead?

Ans: 6.05 cal/g (or 25.2 kJ/kg)

- 49 -

7. a. Inside the sun, two 2H atoms fuse into a

4He atom. The mass

of 2H is 3.3434 x 10

-27kg and

4He is 6.6443 x 10

-27kg. How much energy is released by this

reaction?

Ans: 3.82 x 10-12

J

b. A toy gun uses a spring to throw a plastic 6.00 gram bullet at 20 m/s. How much greater is

the mass of the gun when the spring is compressed than when it isn’t? (Assume 100%

efficiency in transferring the spring’s energy to the bullet.)

Ans: 1.34 x 10-17

kg

8. Short answer, 5 points each:

a. A big dog and a little dog are pulling on opposite ends of a bone. The big dog is winning,

dragging the little dog along the ground. The force on the little dog from the big one is

________ the force on the big dog from the little one. (greater than, less than, or equal to.)

b. The pressure in a cylinder increases from 100 kPa to 150 kPa, due to a change in

temperature. The piston does not move, keeping the volume constant at .002 m3. How much

work is done?

c. The input piston of a hydraulic press has an area of 2 in2 and the output piston has an area of

10 in2. If a 50 lb force pushes on the input piston, what is the pressure on the output piston?

d. Two rocks are thrown from a cliff. One is thrown upward at 10 m/s, the other is thrown

downward at 10 m/s. Just after being thrown, how do the accelerations of these rocks

compare?

e. The belt does not slip. The tangential velocity of pulley B is

________ that of pulley A. (greater than, less than, or equal to.)