- 30 Phy 121
Section 9: Torque & Rotational Dynamics
e
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=lF
Problem 9 – 1: Find the torque in each position:
Weight of an object:
- 31 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
Στ = Iα
Rotational counterpart of Newton's second law.
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 Nm 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
x (or s) θ
v
ω
a
α
F
τ
m
I
KET
KER
examples:
2
2
vf = vi + 2aΔx
2
2
ωf = ωi + 2αΔθ
2
KET = 1/2 mv
2
KER = 1/2 Iω
- 32 Section 10: Statics (Objects at Rest):
⃑
⃑
(or ΣFx = 0 and ΣFy = 0)
1st condition of equilibrium:
(F = ma with a = 0)
Στ = 0
2nd condition of equilibrium:
(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
- 33 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 Nm + 800Nm = (TA)(4m)
1200 Nm = (TA)(4m)
= TA
- 34 TA= 300 N
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
- 35 Section 11: Fluids. The Gas Law.
Density:
ρ=
m = mass, V = volume
Pressure:
P=
F = force, A = area
(ρ = Greek "rho")
(PSI = lb/in2, Pascal = N/m2)
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):
Assume steady, non-turbulent flow of an incompressible, nonviscous fluid. (Viscosity = internal
friction.) Comparing two points in a pipe,
Equation of Continuity:
Bernoulli's Equation:
A1v1 = A2v2
or, Av = constant
2
2
P1 + ½ρv1 + ρgh1 = P2 + ½ρv2 + ρgh2
2
or, P + ½ρv + 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.)
Problem 11-1: The pipe has the same diameter at points A and B,
- 36 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.
Example11-c: A boat displaces 40 m3 of water. What is the buoyant force on it?
Ans: 392 kN
Problem 11-4: A 60 lb bag of cement is loaded into a boat, making it sit lower in the water. How
much water is displaced?
Problem 11-5: A 100 cm3 rock, weighing 3.5 N on dry land, is weighed underwater. What does the
scale read?
Ans: 2.52 N
Problem 11-6: Compare buoyant force in previous problem to buoyant force on a 7.0 N object with
the same volume.
(The fluid doesn't have to be water. Displacing air also gives a buoyant force, which is what makes
- 37 balloons and blimps float.)
Temperature:
Water freezes: 0C, Water boils: 100C.
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 = -273C
Kelvin scale: 0 K = absolute zero
273 K = 0C, 373 K = 100C, and so on.
The Ideal Gas Law:
PV = nRT
V = volume
T = ABSOLUTE temperature
P = ABSOLUTE pressure (from collisions with gas molecules)
23
n = number of moles of gas (1 mole = 6.02 x 10 molecules.)
R = 8.314 J/moleK
Example 11-d: You check your tire pressure at -7C, and the gauge reads 32 psi. If the tire is
heated to 38C, what will the gauge read then?
Ans: 39.9 psi
Problem 11-7: Short answer questions:
a. The volume of an ideal gas is reduced while its pressure is held constant. Does its
temperature increase, decrease, or stay the same?
b. If the temperature is kept constant and V is reduced, what happens to P?
c. If V is kept constant, what happens to P if T is reduced?
Problem 11-8: 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.)
- 38 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 35C. Find its length at -20C.
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 1F.
(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 :
 1cal 
(100 J )

 4.186 J 
=
23.9 cal
Specific Heat Capacity (c) = amount of heat needed to raise 1 gram of the substance 1C.
- 39 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 20C. Find its final temperature.
Ans: 37.9C
By conservation of energy, in an insulated container
ΣQ = 0
Example 12-c: 300 g of lead at 95C is placed in 100 g of water at 15C. What is their common
final temperature?
Ans: 21.7C
Problem 12-2: 90 grams of aluminum at 10C is placed in 200 grams of mercury at 120C. What is
their final equilibrium temperature?
Ans: 37.8C
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 120C is cooled, becoming 50C water. How much heat
does it give off?
Ans: 2.39 kcal
Example 12-e: 4.0 g of steam at 120C is mixed with 70 g of 20C water. What is the mixture's final
temperature? (Assume all steam ends up as liquid.)
Ans: 54.0C
Problem 12-3: 50 g of ice at -30 is heated into 40C water. How much heat did it absorb?
Ans: 6700 cal
Problem 12-4: 50 g of ice at - 30C is placed in contact with 400 g of aluminum at 90C. What is
their final temperature? (Assume all ice ends up as liquid.)
Ans: 22.7C
- 40 Review of sections 9 - 12:
1. At 20C, 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: 334C
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 Nm 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 1F 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?
- 41 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
- 42 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
moleC
or
J _
moleC
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 100C 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:
γ
PV = 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.)
- 43 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.
- 44 Prob. 13-5: Find the net work per cycle done by
this engine.
Ans: 4000 J
Efficiency = work out
energy in
e = Wout
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 =
(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 250C and the exhaust leaves at 105C.
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%
- 45 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
Lorentz - Fitzgerald Contraction:
where
γ=
√
( )
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
- 46 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 - mc2
(KE = 1/2 mv2 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
2
joules. A kilogram, then, is just another unit of energy; mass and energy are equivalent: E = mc
(Applies to all forms of mass and energy.)
2
Eo = mc
= Rest energy (Energy equivalent of the rest mass.)
2
Etot = γmc
= Total energy
KE = Etot - Eo
.
-19
In atomic physics, energy is often measured in electron-volts rather than joules. 1 eV = 1.6 x 10
6
1 MeV = 10 eV.
J.
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 105eV, 3.41 x 105eV
approaches m/0 = . Giving a
Why no object can reach c: As v approaches c, γ m =
√
( )
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 ( ̅ ≈ 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 10-13J, .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
2
1H 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
- 48 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.
a. How much work was done by the engine?
b. How much energy was put out by the
engine?
c. What was the average power?
d. What was the average force on the car?
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, 327C, is poured into 300 g of water at 23.5C, in
a sealed, insulated container. When the lead and water reach thermal equilibrium, they are at
33.5C. What is the heat of fusion of lead?
Ans: 6.05 cal/g (or 25.2 kJ/kg)
- 49 2
4
7. a. Inside the sun, two H atoms fuse into a He atom. The mass
2
-27
4
-27
of H is 3.3434 x 10 kg and He is 6.6443 x 10 kg. 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.)