Download Exam 4 Study Guide - RIT

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Exam 4
Wednesday, 21 May 2008
Room BLDG 7 – 1350……Webb Auditorium
TIME: 2:45 TO 4:45 PM
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NOTE: THIS STUDY GUIDE WILL NOT BE COMPLETELY
FINISHED UNTIL SUNDAY, MAY 18, 2008, 5 PM.
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******BE SURE TO SCROLL ALL THE WAY TO END!!!
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Formula sheet will be provided … see below…
Bring ruler with millimeter markings
Bring a protractor (unlikely, but just in case)
Bring a fully charged calculator
******BE SURE TO SCROLL ALL THE WAY TO END!!!
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Descriptions of some of the topics covered

Unit conversion (Conversion relations will be given)

Scientific notation

Significant figures and rules for arithmetic operations

Measurement uncertainty

Proper form
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CONTENT OF EXAM:
About 50% MULTIPLE CHOICE ON THE WHOLE COURSE
About 50% MOMENTUM AND ENERGY CONSERVATION
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Preliminary
Formula sheet (preliminary)
Unit Systems
Length Mass Time Force
SI / MKS
m
kg
s
N
CGS
cm
g
s
dyne
BE
ft
slug s
lb

Significant figures
 In Dr. Entenberg’s class, trailing zeroes will be meaningful unless the number has quotes
around it, i.e. “75,000” nails.
 Use a decimal point when you know the correct number of significant figures.
 Whole numbers have no decimal point; e.g., 3 = 3.0000000…….
 If unsure, use scientific notation to determine the number of significant figures.
 Proper form: [(A + A) 10exponent units] where A must be written to 1 significant figure
Note that A and A must have
1. Common exponent outside the parentheses
2. Common units outside the parentheses
3. Same number of decimal places inside the parentheses
Scientific notation example: 3769. m = 3.769 X 103 m
Uncertainty propagation examples:

f=X*Y
f = fmax – f where fmax = (X + X) * (Y + Y)

f=X/Y
f = fmax – f where fmax = (X + X) / (Y – Y)
Constants
Geometry
Trigonometry
g = 9.80 m/s2 g = 980. cm/s2 g = 32.2 ft/s2 G = 6.67 x 10-11 N m2 / kg2
2  R  R2 4  R2 (4/3)  R3  R2 H
sine = opp / hyp
cosine = adj / hyp tangent = opp / adj = sine/cosine
Vectors in 2-D
Vectors in 2-D
Vectors in 2-D
Symbol
F = |F| = magnitude ( + ONLY)
Fx = component along x direction ( + )
Fx= + F cos  (or Fx= + F sin ’)
Force and Vector NOTATION:

    
Fnet   F  F1  F2  F3  F4   =
 = angle wrt a given direction ( + )
Fy = component along y direction ( + )
Fy= + F sin  (or Fy= + F cos ’)
Fnet =  F = F1 + F2 + F3 + F4….
Newton’s Second Law Notation
Fnet =  F = F1 + F2 + F3 + …. = M a = 0 (Vector notation form)
Fnet, x =  Fx = F1x + F2x + F3x + ….. = M ax = 0 (Component form, more practical)
Fnet, y =  Fy = F1y + F2y + F3y + ….. = M ay = 0 (Component form, more practical)
Force: Newton’s Third Law All forces occur in “action-reaction” pairs! There is no such thing as an
isolated force!
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Variables, definitions, and equations
Mass
Linear variables
X and Y motions
“linear mass”
m or M
Angular variables
Tangential variables
Rigid body rotation
Point P on rigid body
“moment of inertia” angular mass
I =  mrperp2 (see below for more info)
Position
Displacement
Average velocity
Average acceleration
Linear variables
X
X = Xfinal – Xinitial
vxavg = x / t
axavg = vx / t
Angular variables



 =  final –  initial
avg =  t
avg =  t
Constant acceleration equations
Velocity time equation
Position time equation
Velocity position equation
aX = (aX)avg = constant
Vxf = Vxi + ax t
Xf = Vxi t + ½ ax t2
Vxf 2 = Vxi 2 + 2 ax X
 = avg = constant
f t
fi t + ½  t2
f 2 = i 2 + 2 
Force: Newton’s 2nd Law
Centripetal acceleration
 Fx = M ax
ac = V2 / r
Torque: Newton’s 2nd Law …………………………………………………………… = I 
Work
Energy conservation
Momentum
Momentum conservation
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W = |F| cos |s|
Ef = Ei
Etotal = K + U
(no friction)
Etotal = K + U + Ethermal (friction)
K = ½ M V2
Ugravity = M g Y
Uspring = ½ k X2
px = M vx
M1V1i + M2V2i = M1V1i + M2V2i


 = + r F +CCL , -CL
 = + r F sin()
K = ½ I 2
L=I
Isystem i system i = Isystem f system f
I1i1i + I2i2i = I1f1f + I2f2f
Tangential variables
Stan = r 


Vtan = r 
atan = r 
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Various Force Magnitudes
Gravitational force (general)
Gravity force at earth’s surface
Friction force magnitude
Normal force magnitude
Tension force magnitude
Spring force magnitude
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F = G M1 M2 / r2
G = 6.67 x 10-11 N m2 / kg2
Wt = M g where g = [G ME / RE2] = 9.80 m/s2 
fs < fsmax fsmax =s Fn fkin = k Fn
Fn or n
Ft or T
|F| = k |x|
Two dimensional circular motion: Centripetal (radial) acceleration = ar = V2 / r = 2 r
Center of mass (center of gravity)
{Newton’s Universal Law of Gravitation}
Tangential acceleration = at = r 
Xcg = (M1 X1 + M2 X2 + … + MN XN) / (M1 + M2 + … + MN)
Moment of inertia: point mass (m r2), hoop (M R2), solid disk (½ M R2), sphere [(2/5) M R2 ], rod, at end [(1/3) M L2 ]
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
You are given the Earth with its mass of 5.98 x 1024 kg and its radius of 6.37 x
106 m. You are also given that g = 9.80 m /s2 and that G = 6.67 X 10-11 N m2 / kg2
(a) Use the standard method to calculate the weight of a 70.0 kg person standing
on the Earth’s surface. Use SI units and 3 significant figures. (686. N)
(b) Use Newton’s Law of Universal Gravitation to calculate the weight of the same
70.0 kg person. Start by making a cartoon sketch of the person standing on the
surface of the earth; sketch and label the radius of the Earth on your drawing.
Use SI units and 3 significant figures. (686. N)
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IN CLASS EXAMPLE
y axis
T
x axis
60
 A man pulls on a 70.0 kg crate with a rope. The rope makes an angle of 60.0 with
respect to the vertical direction. The tension in the rope is 450. N. As the crate slides,
its motion is opposed by a kinetic friction force of 140. N. The crate is displaced to right
a distance of 8.00 m.
(a) Sketch and label all the forces that act on the crate. Place the tail of the vector at
the approximate point where the force acts.
(b) Calculate the work by the man on the crate. 3.12 x 103 J
(c) Calculate the work by the force of friction on the crate. – 1.12 x 103 J
(d) Calculate the work done by any other forces on the crate. Normal 0, Gravity 0
(e) Calculate the kinetic energy of the crate after it has moved the distance of 8.00
m. 2.00 x 103 J
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
EOC Problem 50 modified… You are given a child on a swing. If you wish,
you can assume that the mass of the child is 35.0 kg. She is at the top of her
motion when the angle is 45.0°.
(a) Let the zero for gravitational potential energy be at the bottom of her swing.
Show with a drawing that the Y coordinate for the potential energy function
M g Y is Y = L ( 1 – cos( ) ) where  is the angle between the vertical
and the rope of the swing and L is the length of the rope.
(b) Starting with Efinal = Einitial, find her speed at the bottom of the motion.
(4.15 m/s)
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Total Kinetic Energy
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Ktotal = Ktranslational + Krotational
 Calculate the total kinetic energy of a bowling ball of mass M, radius R, and
speed V. Assume that the bowling ball is a sphere with a uniform distribution
of mass; ignore the holes. The ball rolls without slipping on a smooth level
surface. Be sure to show the formulas and the numerical setup.
Let M = 7.00 kg, R = 11.0 cm, and V = 3.00 m/s.
Ktrans = ½ M V2 = ½ (7.00 kg) (3.00 m/s)2 = 31.5 Joules
Krot
Krot
= ½ I 2 = ½ ( 2/5 M R2 ) ( V / R )2
= ½ I 2 = ½ ( 2/5 7.00 kg (0.110 m)2 ) ( 3.00 m/s / 0.110 m )2 = 12.6 Joules
Ktotal = 44.1 Joules
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 Use energy conservation to show that a circular object of mass M and moment of
inertia I , that rolls down a ramp of height h without slipping has the following
final speed.
Vfinal = [ 2 g h / ( 1 + ( I / M R2) ] ½
Explain which object (hoop, cylindrical disk, or sphere) with mass M and radius R
will reach the bottom first?
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 Let the mass of the solid circular disk be M = 7.00 kg and let m = 3.50 kg. The
radius of the mass-less pulley is Rp = 0.150 meter and Rdisk = 0.300 meter. Let
the system start from rest. There is no friction.
 Find the following quantities after the mass m
has gone down by 20.0 m.
RMS with aluminum
disk attached
(a) Find the speed V of the mass m
Efinal = Einitial
½ m Vf2 + ½ I f2 + m g Yf = ½ m Vi2 + ½ I i2 + m g Yi
½ 3.50 Vf2 + ½ ( ½ 7.00 Rd2 ) (Vf / Rp)2 + 0
= 0 + 0 + 3.50 9.80 20.0
2
2
½ 3.50 Vf + ½ ( 14.0 Vf ) + 0 = 0 + 0 + 686.
Vf = 8.85 m/s
(b) Find the angular velocity of the disk
f = Vf / Rp = 59.0 radians/s
(c) Find the total energy of the system 686. Joules
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R
m
Figure 1 Mass/pulley system
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Example 10.8 Thermal energy
created sledding down a hill.
(a) George jumps onto his sled and
starts from rest at the top of a 5.00 m
high hill. His speed at the bottom is
8.00 m/s. The mass of George and
the sled is 55.0 kg. How much
thermal energy was produced in the
process?

E = K + U + Ethermal
Note: System = (sled and rider) + surface of slope + earth
Ei = Ef
Ki + Ui + Eth,i = Kf + Uf + Eth,f
Ki + Ui = Kf + Uf + (Eth,f – Eth,i)
Ki + Ui = Kf + Uf + Ethermal
Note: Ethermal = (Eth,f – Eth,i)
½ M Vi2 + M g Yi = ½ M Vf2 + M g Yf + Ethermal
0 + 55.0 · 9.80 · 5.00 = ½ 55.0 · 8.002 + 0 + Ethermal
Ethermal = 935. J
(b) Suppose that the length of the hill is 12.0 m long. What is the average kinetic
friction force on the sled as it moves down the hill? (77.9 N) Explain your
reasoning for this calculation.
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Problem 62 A new event for the Winter Olympics. An athlete will sprint 100. m, starting from rest, then
leap onto a 20.0 kg bobsled. The person and bobsled will then slide down a 50.0 m long ice-covered
ramp, sloped at 20.0°, and into a spring with a carefully calibrated spring constant of 2000. N/m. Lisa,
whose mass is 40.0 kg, can reach a speed of 12.0 m/s in the 100. m dash. Ignore friction on the icecovered ramp!
(a) How fast will Lisa and the sled be going when they start to go down the ramp?
There will be an inelastic collision between Lisa and the sled.
Energy will be lost because the friction between Lisa and the sled.
Momentum will be conserved because the sled will slide without friction on the ice.
M1 V1i + M2 V2i = M1 V1f + M2 V2f
V1f = V2f = Vf = 8.00 m/s
(b) How fast will Lisa and the sled be going when they first hit the spring?
Energy will be conserved as Lisa and the sled go down the frictionless ramp which is 17.1 m high.
Momentum will not be conserved because Lisa and the sled are under the influence of gravity.
Vtop = 8.00 m/s Vbottom = 20.0 m/s
(c) How far will Lisa and the sled compress the spring?
Energy will be conserved as Lisa and the sled compress the spring whose potential energy is U = ½ k X2.
Momentum will not be conserved because Lisa and sled are interacting with the spring.
½ M Vf2 + ½ k Xf2 = ½ M Vi2 + ½ k Xi2 M = M1 + M2 = 60.0 kg, Vi = Vbottom = 20.0 m/s, Xi = 0
Xf = 3.46 m
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Example 10.11 Energy Transformations in a perfectly inelastic collision
Show that Ethermal = 1.84 J
Only momentum is conserved in this inelastic interaction
M1 V1i + M2 V2i = M1 V1f + M2 V2f
V1f = V2f = Vf
0.200  3.00 + 0.400  (-2.25) = 0.200 Vf + 0.400 Vf
0.600 + ( – 0.900) = 0.600 Vf
Vf = – 0.500 m/s
The energy lost goes into thermal energy
E = Ksystem + U + Ethermal
U = constant = 0 because level stays the same.
Ei = Ef
Ksystemi + Eth,i = Ksystemf + Eth,f
Ksystemi = Ksystemf + Ethermal
Note: Ethermal = (Eth,f – Eth,i)
½ 0.200 3.002 + ½ 0.400 (-2.25)2 = ½ 0.200 (-.500)2 + ½ 0.400 (-.500)2 + Ethermal
0.900 + 1.013 = 0.025 + 0.050 + Ethermal
Ethermal = 1.84 J
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Example from class…Velocities in an elastic collision
M1
M2
x-axis

Given a one dimensional elastic collision between a ball and a block of wood at
rest. The mass of the ball is 8.00 kg and the mass of the wood is 16.0 kg. The initial
speed of the ball is 45.0 m/s.
(a) Using the appropriate equations, determine the final x-component of velocity
(V1f) for the ball (M1) and the final x-component of velocity (V2f) for block of wood
(M2). (-15.0 m/s, +30.0 m/s)
(b) Show with a calculation (to 3 significant figures) that momentum was
conserved in this collision. 360. kg m/s = 360 kg m/s
(c) Show with a calculation (to 3 significant figures) that this collision was elastic.
(8100. Joules = 8100. Joules)
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Average Power =
Average Power =
Units
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Work / Time
Fx Vx
[ Power ] = [ Work ] / [ Time ] = Joules / Second = Watt
Problem 10 – 38
(a) How much work does an elevator motor do to lift a 1000. kg elevator a
height of 100. m? 9.80 x 105 Joules
(b) How much power must the motor supply to do this in 50.0 s at constant
speed? 1.96 x 104 Watts
Problem 10 – 44
A 710. kg car drives at a constant speed of 23.0 m/s. It is subject to a drag force
of 500. N. What power is required from the car’s engine to the drive the car
(a) on level ground? 1.15 x 104 Watts
(b) up a hill with a slope of 2.00°. (optional)
Example from class
How much does it cost to operate a 100. Watt bulb for 1 day? Assume that a
“kilowatt hour” (kwh) costs 15.0 cents. $0.36
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