Physics 201 – Lecture 23
Lecture 23
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Basic Concepts and Quantities
Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room
CH Sections 604 605 606 609 610 611 (TA: Moriah T., Abdallah
C., Tamara S.)
VV Sections 602 603 607 608 612 (TA: Eric P., Dan C., Zhe D.)
Format: Closed book, three 8 x11” sheets, hand written
Electronics: Any calculator is okay but no web/cell access
Quiet room: Test anxiety, special accommodations, etc.
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Chapters Covered
Chapter 9: Linear momentum and collision (not 9.9)
Chapter 10: Rotation about fixed axis and rolling
Chapter 11: Angular Momentum (not 11.5)
Chapter 12: Static equilibrium and elasticity
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Momentum, Angular Momentum
v Linear Momentum, Impulse
v Angular Momentum (magnitude and direction)
Torque
Collisions: Elastic & Inelastic
Center of Mass
Rotational Motion (1-axis)
Angular displacement (•θ) / Velocity(ω), Acceleration (α).
Moments of Inertia
Rotational Kinetic Energy
Conservation Laws: Energy, Momentum, Angular Momentum
Static Equilibrium
Elasticity: Young’s, Shear, Bulk Modulus
Physics 201: Lecture 23, Pg 1
Physics 201: Lecture 23, Pg 2
Chapter 9
Chapter 9 : Momentum and Momentum Conservation
Ix =
Ix
Physics 201: Lecture 23, Pg 3
Physics 201: Lecture 23, Pg 4
Chapter 10
Chapter 9
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At a point a distance R away from the axis of rotation, the
tangential motion:
v=ωR
Center of Mass
= θR
v vT (tangential) = ω R
v aT
=αR
v s (arc)
N
r
r
r
r
∑ m i ri
m r + m 2 r2 + m 3 r3 + L
r
i =1
rCM ≡ N
= 11
M
∑ mi
i =1
r
∑ m i vi
N
r
v CM ≡
i =1
N
∑
i =1
mi
α = constant
r
r
r
m v + m 2 v 2 + m 3v3 + L
= 1 1
M
Physics 201: Lecture 23, Pg 5
s
ω
(angular accelation in rad/s 2 )
α
ω f = ωi + αt (angular velocity in rad/s)
v = rω
1
θ f = θ i + ωi t + α t 2
2
at = rα
ω 2f = ωi2 + 2α (θ f − θ i )
θ
R
ac =
v2
= ω 2r
r
Physics 201: Lecture 23, Pg 6
Page 1
Physics 201 – Lecture 23
Chapter 10
Chapter 10
Physics 201: Lecture 23, Pg 7
Physics 201: Lecture 23, Pg 8
Chapter 12
Chapter 11
r
r dL
∑τ =
dt
Physics 201: Lecture 23, Pg 9
Physics 201: Lecture 23, Pg 10
Approach to Statics:
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In general, we can use the two equations
r
∑τ = 0
r
∑F = 0
Changes in length: Young’s modulus
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Young’s modulus: measures the resistance of a solid
to a change in its length.
F
to solve any statics problems.
elasticity in length
A
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L0
When choosing axes about which to calculate torque,
choose one that makes the problem easy....
However if there is acceleration, then restrict rotation axis
to center of mass (as well as for translation).
∆L
Y = tensile stress =
tensile strain
Physics 201: Lecture 23, Pg 11
F/A
∆L / L0
Physics 201: Lecture 23, Pg 12
Page 2
Physics 201 – Lecture 23
Changes in shape: Shear Modulus
Changes in volume: Bulk Modulus
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Applying a force perpendicular to a surface
Bulk modulus: measures the resistance of solids or
liquids to changes in their volume.
S=
shear stress F / A
=
shear strain ∆x / h
Volume elasticity
V0
B=−
F/A
•V/V i
F
Vi + ∆V
P ≡ F/A = Pressure
Physics 201: Lecture 23, Pg 13
Physics 201: Lecture 23, Pg 14
Comparison:
Comparison
Kinematics
Angular
Angular
α = constant
ω = ω 0 + αt
Linear
I = Σi mi ri2
Linear
a = constant
τ=rxF=αI
v = v 0 + at
L=rxτ=
Iω
m
F=am
p = mv
1
θ = θ 0 + ω 0t + α t 2
2
x = x0 + v 0t + 12 at
ω 2 − ω 0 = 2αθ
v 2 − v 0 = 2 ax
W = τ ∆θ
v AVE = 12 ( v + v 0 )
1
K = Iω 2
2
K=
∆K = WNET
∆K = WNET
2
ωAVE = 12 (ω + ω0 )
τ EXT
2
2
dL
=
dt
Physics 201: Lecture 23, Pg 15
FEXT =
dp
dt
W = F •∆x
1 2
mv
2
Physics 201: Lecture 23, Pg 16
Momentum and collisions
Momentum and collisions
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Remember vector components
mvi cos θ
mvi
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Remember vector components
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A 5 kg cart rolling without friction to the right at 10 m/s
collides and sticks to a 5 kg motionless block on a 30°
frictionless incline.
How far along the incline do the joined blocks slide?
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Dynamics
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A 5 kg cart rolling without friction to the right at 10 m/s collides
and sticks to a 5 kg motionless block on a θ=30° frictionless
incline.
Momentum parallel to incline is conserved
Normal force (by ground on cart) is ⊥ to the incline
mvi cos θ + m 0 = 2m vf à vf = vi cos θ / 2 = 4.4 m/s
Now use work-energy
2mgh + 0 = ½ 2m v2f à d = h / sin θ = v2f / (2g sin θ)
Physics 201: Lecture 23, Pg 17
Physics 201: Lecture 23, Pg 18
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Physics 201 – Lecture 23
Example problem: Going in circles A
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Example problem: Going in circles B
A 2.0 kg disk tied to a 0.50 m string undergoes circular motion
on a rough but horizontal table top. The kinetic coefficient of
friction is 0.25. If the disk starts out at 5.0 rev/sec how many
revolutions does it make before it comes to rest?
Work-energy theorem
W = F d = 0 – ½ mv2
F = -µmg d = - ½ mv2
d= v2 /(2µg)=(5.0 x 2π x 0.50)2/ (0.50 x 10) m = 5 π2 m
Rev = d / 2πr = 16 revolutions
What if the disk were tilted by 60° ?
A 2.0 kg disk tied to a long string undergoes circular motion on
a frictionless horizontal table top. The string passes through a
hole and then hangs vertically. The disk starts out at 5.0
rev/sec 0.50 m away from hole. If you pull slowly down on the
string so that the final radius is 0.25 m, what is the final
angular velocity?
No external torque so angular momentum is conserved.
Ib ωb = Ia ωa
I = mR2
(0.50)2 ωb = (0.25)2 ωa
ri
4 ωb = ωa = 20 rev/sec
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ωi
Top
view
Physics 201: Lecture 23, Pg 19
Physics 201: Lecture 23, Pg 20
Spinning ball on incline
Spinning ball on incline
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A solid disk of mass m and radius R is spinning with
angular velocity ω. It is positioned so that it can either move
directly up or down an incline of angle θ (but it is not rolling
motion). The coefficient of kinetic friction is µ. At what
angle will the disk’s position on the incline not change?
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A solid disk of mass m and radius R is spinning with
angular velocity ω. It is positioned so that it can either move
directly up or down an incline of angle θ (but it is not rolling
motion). The coefficient of kinetic friction is µ.
At what angle will the disk’s position on the incline not
change?
N
r
∑F
ω
∑ Fx
=0
= 0 = f − mg sin θ
= 0 = N − mg cos θ
f = µN = µmg cos θ
∑ Fy
θ
µ = tanθ
Physics 201: Lecture 23, Pg 22
Example Problem
Spinning ball on incline
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A solid disk of mass m and radius r is spinning with angular
velocity ω. It is positioned so that it can either move directly up
or down an incline of angle θ (but it is not rolling motion). The
coefficient of kinetic friction is µ. While spinning the disk’s
position will not change.
How long will it be before it starts to roll?
N
This will occur only when ω=0.
∑τ z ≠ 0 about disk CM
o
∑ τ z =| r || f | sin φ = rf sin 90
Iα = rf = rµmg cos θ
α = 2 µg cos θ / r
θ
mg
Physics 201: Lecture 23, Pg 21
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f
ω
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f
ω
A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The
rod is vertical and is anchored to the ground through a
frictionless pivot. It sits perfectly balanced at an unstable
equilibrium. A 500 gm bullet is shot horizontally at 100 m/s
through the mass. The force versus time plot is shown.
How fast is the bullet going when
it leaves?
What is the tension in the rod just
after the bullet exits?
Method: Impulse
I = area under curve
θ
I x = ∫ Fx (t )dt
mg
I x = ∫ [ 225 − 10 4 (t − 0.15) 2 ]dt
ω f = 0 = −ω + α t = −ω + 2tµg cos θ / r
Physics 201: Lecture 23, Pg 23
Physics 201: Lecture 23, Pg 24
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Physics 201 – Lecture 23
Example Problem
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Example Problem
A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The
rod is vertical and is anchored to the ground through a
frictionless pivot. It sits perfectly balanced at an unstable
equilibrium. A 500 gm bullet is shot horizontally at 100 m/s
through the mass. The force versus time plot is shown.
How fast is the bullet going when
it leaves?
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I x = ∫ [ 225 − 10 4 (t − 0.15) 2 ]dt
Pf′ = 45 Ns
I x = 225t − 10 (t − 0.15) / 3 |
4
3
0.3
0.0
v ′f = 45 / 3 = 15 m/s
v2
= − mg − T
∑ Fy = ma c = − m
r
T = [3(15 2 ) / 1.5 − 30 ]N
I x = 45 Ns
Pf = Pi − 45 Ns
= (0.5 × 100 − 45 ) Ns
Pf = 5 Ns v f = 10 m/s
A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The
rod is vertical and is anchored to the ground through a
frictionless pivot. It sits perfectly balanced at an unstable
equilibrium. A 500 gm bullet is shot horizontally at 100 m/s
through the mass. The force versus time plot is shown.
What is the tension in the rod just
after the bullet exits?
T = (450 − 30) N = 420 N
Physics 201: Lecture 23, Pg 25
Statics: Example
Physics 201: Lecture 23, Pg 26
Statics: Example
A sign of mass M is hung 1 m from the end of a 4 m long
beam (mass m) as shown in the diagram. The beam is hinged
at the wall. What is the tension in the wire in terms of m, M, g
and any other given quantity?
T
30°
Fy
X
Fx
2m
mg
3m
Mg
Process: Make a FBD and note known / unknown forces.
wire
θ = 30ο
Chose axis of rotation at support because Fx & Fy are not known
Σ F = 0 à 0 = Fx – T cos 30°
1m
0 = Fy + T sin 30° - mg - Mg
z-dir: Στz = 0 = -mg 2r – Mg 3r + T sin 30° 4r (r = 1m)
SIGN
The torque equation get us where we need to go, T
T = (2m + 3M) g / 2
Physics 201: Lecture 23, Pg 27
Physics 201: Lecture 23, Pg 28
Center of Mass Example: Astronauts & Rope
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(1)
Two astronauts are initially at rest in outer space and 20
meters apart. The one on the right has 1.5 times the
mass of the other (as shown). The 1.5 m astronaut
wants to get back to the ship but his jet pack is broken.
There happens to be a rope connected between the
two. The heavier astronaut starts pulling in the rope.
(1) Does he/she get back to the ship ?
(2) Does he/she meet the other astronaut ?
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Example: Astronauts & Rope
There is no external force so if the larger astronaut
pulls on the rope he will create an impulse that
accelerates him/her to the left and the small
astronaut to the right. The larger one’s velocity will
be less than the smaller one’s so he/she doesn’t let
go of the rope they will either collide (elastically or
inelastically) and thus never make it.
m
M = 1.5m
Physics 201: Lecture 23, Pg 29
M = 1.5m
Physics 201: Lecture 23, Pg 30
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Physics 201 – Lecture 23
Example: Astronauts & Rope
Forces and rigid body rotation
(2) However if the larger astronaut lets go of the rope
he will get to the ship. (Too bad for the smaller
astronaut!)
In all cases the center of mass will remain fixed
because they are initially at rest and there is no
external force.
To find the position where they meet all we need do is
find the Center of Mass
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To change the angular velocity of a rotating object, a force
must be applied
How effective an applied force is at changing the rotation
depends on several factors
v The magnitude of the force
v Where, relative to the axis of rotation the force is applied
v The direction of the force
r
r
F
F
M = 1.5m
A
B
C
Which applied force will cause the wheel to spin the fastest?
Physics 201: Lecture 23, Pg 31
Physics 201: Lecture 23, Pg 32
More on torques
Leverage
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The same concept applies to leverage
vthe lever undergoes rigid body rotation about a pivot
point:
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r
F
B
C
A
r
F
r
F
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Which applied force provides
the greatest lift ?
You need to change the tire on
you car. You use a tire wrench
which allows you to apply a pair
of forces.
(A) What is the torque produced
by a tire wrench of length L,
given an applied couple of
magnitude F, acting on a lug
nut (point F) as shown in the
figure?
(B) Assume the lug nut is stuck
What is the torque acting on the
wheel, if the lug nut is a
distance r from the center?
Image
courtesy
John Wiley
& Sons, Inc.
Physics 201: Lecture 23, Pg 33
Physics 201: Lecture 23, Pg 34
For Thursday
Wheel wrench
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1. τF = (L/2) F + (L/2) F = LF
φ
Chapter 13 (Newton’s Law of Gravitation)
φ
2. τF = r F sinφ + r F sin ( π−θ )=
= LF
3. τF = L F + 0 F
Notice the torque is the same
everywhere.
Physics 201: Lecture 23, Pg 35
Physics 201: Lecture 23, Pg 36
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Physics 201 – Lecture 23
Momentum & Impulse
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Momentum & Impulse
A rubber ball collides head on (i.e., velocities are
opposite) with a clay ball of the same mass. The balls
have the same speed, v, before the collision, and stick
together after the collision. What is their speed
immediately after the collision?
A.
B.
C.
D.
A rubber ball collides head on with a clay ball of the
same mass. The balls have the same speed, v, before
the collision, and stick together after the collision. What
is their speed after the collision?
(a) 0
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0
½v
2v
4v
(b) ½ v
(c) 2 v
(d) 4 v
Physics 201: Lecture 23, Pg 37
Physics 201: Lecture 23, Pg 38
Momentum, Work and Energy
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Momentum, Work and Energy
A 0.40 kg block is pushed up against a spring (with spring
constant 270 N/m ) on a frictionless surface so that the spring is
compressed 0.20 m. When the block is released, it slides across
the surface and collides with the 0.60 kg bob of a pendulum.
The bob is made of clay and the block sticks to it. The length of
the pendulum is 0.80 m. (See the diagram.)
To what maximum height above the surface will the ball/block
assembly rise after the collision? (g=9.8 m/s2)
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A. 2.2 cm
B. 4.4 cm
C. 11. cm
D. 22 cm
E. 44 cm
F. 55 cm
A. 2.2 cm
B. 4.4 cm
C. 11. cm
D. 22 cm
E. 44 cm
F. 55 cm
Physics 201: Lecture 23, Pg 39
Physics 201: Lecture 23, Pg 40
Momentum, Work and Energy ( Now with friction)
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A 0.40 kg block is pushed up against a spring (with spring
constant 270 N/m ) on a frictionless surface so that the spring is
compressed 0.20 m. When the block is released, it slides across
the surface and collides with the 0.60 kg bob of a pendulum. The
bob is made of clay and the block sticks to it. The length of the
pendulum is .80 m. (See the diagram.)
To what maximum height above the surface will the ball/block
assembly rise after the collision?
A 0.40 kg block is pushed up against a spring (with spring
constant 270 N/m ) on a surface.
If mstatic = 0.54, how far can the spring be compressed and the
block remain stationary (i.e., maximum static friction)?
ΣF=0= ku-f=ku-µN
u = µ mg/k = 0.54 (0.40x10 N) / 270 N/m= 0.0080 m
Momentum, Work and Energy ( Now with friction)
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Physics 201: Lecture 23, Pg 41
A 0.40 kg block is pushed up against a spring (with spring
constant 270 N/m ) on a surface. The spring is compressed
0.20 m
If mkinetic = 0.50 and the block is 9.8 m away from the
unstretched spring, how high with the clay/block pair rise?
Emech (at collision) = Uspring + Wfriction = ½ k u2 - µ mg d
1/2 m v2 = 135(0.04)-0.50(0.40x10.)10.=(540-20) J=520 J
v2 = 1040/0.40 m2/s2
Now the collision (cons. of momentum) and the swing.
Physics 201: Lecture 23, Pg 42
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Physics 201 – Lecture 23
Momentum and Impulse
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Momentum and Impulse
Henri Lamothe holds the world record for the highest shallow
dive. He belly-flopped from a platform 12.0 m high into a tank
of water just 30.0 cm deep! Assuming that he had a mass of
50.0 kg and that he stopped just as he reached the bottom of
the tank, what is the magnitude of the impulse imparted to
him while in the tank of water (in units of kg m/s)?
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(a) 121
(b) 286
(c) 490
(d) 623
(e) 767
Henri Lamothe holds the world record for the highest
shallow dive. He belly-flopped from a platform 12.0 m high
into a tank of water just 30.0 cm deep! Assuming that he
had a mass of 50.0 kg and that he stopped just as he
reached the bottom of the tank, what is the magnitude of
the impulse imparted to him while in the tank of water (in
units of kg m/s)?
(a) 121
(b) 286
(c) 490
(d) 623
(e) 767
∆p = sqrt(2x9.8x12.3)x50
Physics 201: Lecture 23, Pg 43
Physics 201: Lecture 23, Pg 44
Momentum & Impulse
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Momentum & Impulse
Suppose that in the previous problem, the positively
charged particle is a proton and the negatively charged
particle, an electron. (The mass of a proton is approximately
1,840 times the mass of an electron.) Suppose that they are
released from rest simultaneously. If, after a certain time,
the change in momentum of the proton is ∆p, what is the
magnitude of the change in momentum of the electron?
(a)
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∆p / 1840
Suppose that in the previous problem, the positively
charged particle is a proton and the negatively charged
particle, an electron. (The mass of a proton is approximately
1,840 times the mass of an electron.) Suppose that they are
released from rest simultaneously. If, after a certain time,
the change in momentum of the proton is ∆p, what is the
magnitude of the change in momentum of the electron?
(a)
∆p / 1840
(b) ∆p
(b) ∆p
(c) 1840 ∆p
(c) 1840 ∆p
Physics 201: Lecture 23, Pg 45
Physics 201: Lecture 23, Pg 46
Conservation of Momentum
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Conservation of Momentum
A woman is skating to the right with a speed of 12.0 m/s
when she is hit in the stomach by a giant snowball moving
to the left. The mass of the snowball is 2.00 kg, its speed
is 25.0 m/s and it sticks to the woman's stomach. If the
mass of the woman is 60.0 kg, what is her speed after the
collision?
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A woman is skating to the right with a speed of 12.0 m/s
when she is hit in the stomach by a giant snowball moving
to the left. The mass of the snowball is 2.00 kg, its speed
is 25.0 m/s and it sticks to the woman's stomach. If the
mass of the woman is 60.0 kg, what is her speed after the
collision?
(a) 10.8 m/s
(a) 10.8 m/s
(b) 11.2 m/s
(b) 11.2 m/s
(c) 12.4 m/s
(c) 12.4 m/s
(d) 12.8 m/s
(d) 12.8 m/s
Physics 201: Lecture 23, Pg 47
Physics 201: Lecture 23, Pg 48
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Physics 201 – Lecture 23
Conservation of Momentum
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Momentum and Impulse
Sean is carrying 24 bottles of beer when he slips in a large
frictionless puddle. He slides forwards with a speed of 2.50
m/s towards a very steep cliff. The only way for Sean to
stop before he reaches the edge of the cliff is to throw the
bottles forward at 20.0 m/s (relative to the ground). If the
mass of each bottle is 500 g, and Sean's mass is 72 kg,
what is the minimum number of bottles that he needs to
throw?
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A stunt man jumps from the roof of a tall building, but
no injury occurs because the person lands on a large,
air-filled bag. Which one of the following statements
best describes why no injury occurs?
(a) The bag provides the necessary force to stop the person.
(b) The bag reduces the impulse to the person.
(c) The bag reduces the change in momentum.
(d) The bag decreases the amount of time during which the
momentum is changing and reduces the average force on
the person.
(e) The bag increases the amount of time during which the
momentum is changing and reduces the average force on
the person.
(a) 18
(b) 20
(c) 21
(d) 24
(e) more than 24
Physics 201: Lecture 23, Pg 49
Physics 201: Lecture 23, Pg 50
Momentum and Impulse
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Momentum and Impulse
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A stunt man jumps from the roof of a tall building, but
no injury occurs because the person lands on a large,
air-filled bag. Which one of the following statements
best describes why no injury occurs?
(a) The bag provides the necessary force to stop the person.
(b) The bag reduces the impulse to the person.
(c) The bag reduces the change in momentum.
(d) The bag decreases the amount of time during which the
momentum is changing and reduces the average force on
the person.
(e) The bag increases the amount of time during which the
momentum is changing and reduces the average force on
the person.
Two blocks of mass m1 = M and m2 = 2M are both sliding
towards you on a frictionless surface. The linear momentum
of block 1 is half the linear momentum of block 2. You apply
the same constant force to both objects in order to bring
them to rest. What is the ratio of the two stopping distances
d2/d1?
(a) 1/ 2
(b) 1/ 2½
(c) 1
(d) 2½
(e) 2
(f) Cannot be determined without knowing the masses of
the objects and their velocities.
Physics 201: Lecture 23, Pg 51
Physics 201: Lecture 23, Pg 52
Momentum and Impulse
Momentum and Impulse
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Two blocks of mass m1 = M and m2 = 2M are both sliding
towards you on a frictionless surface. The linear momentum
of block 1 is half the linear momentum of block 2. You apply
the same constant force to both objects in order to bring
them to rest. What is the ratio of the two stopping distances
d2/d1?
In a table-top shuffleboard game, a heavy moving puck collides with a
lighter stationary puck as shown. The incident puck is deflected through
an angle of 20° and both pucks are eventually brought to rest by friction
with the table. The impulse approximation is valid (i.e.,the time of the
collision is short relative to the time of motion so that momentum is
conserved).
Which of the following
statements is correct?
A. The collision must be inelastic because the pucks have different masses.
B. The collision must be inelastic because there is friction between the
pucks and the surface.
C. The collision must be elastic because the pucks bounce off each other.
D. The collision must be elastic because, in the impulse approximation,
momentum is conserved.
E. There is not enough information given to decide whether the collision is
elastic or inelastic.
(a) 1/ 2
(b) 1/ 2½
(c) 1
(d) 2½
(e) 2
(f) Cannot be determined without knowing the masses of
the objects and their velocities.
Physics 201: Lecture 23, Pg 53
Physics 201: Lecture 23, Pg 54
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Physics 201 – Lecture 23
Exercise: Ladder against smooth wall
Momentum and Impulse
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In a table-top shuffleboard game, a heavy moving puck collides with a
lighter stationary puck as shown. The incident puck is deflected through
an angle of 20° and both pucks are eventually brought to rest by friction
with the table. The impulse approximation is valid (i.e.,the time of the
collision is short relative to the time of motion so that momentum is
conserved).
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Which of the following
statements is correct?
A. The collision must be inelastic because the pucks have different masses.
B. The collision must be inelastic because there is friction between the
pucks and the surface.
C. The collision must be elastic because the pucks bounce off each other.
D. The collision must be elastic because, in the impulse approximation,
momentum is conserved.
E. There is not enough information given to decide whether the collision is
elastic or inelastic.
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Bill (mass M) is climbing a ladder (length
L, mass m) that leans against a smooth
wall (no friction between wall and ladder).
A frictional force F between the ladder
and the floor keeps it from slipping. The
angle between the ladder and the wall is
φ.
What is the magnitude of F as a function
of Bill’s distance up the ladder?
φ
L
m
Bill
F
Physics 201: Lecture 23, Pg 55
Physics 201: Lecture 23, Pg 56
Ladder against smooth wall...
Ladder against smooth wall...
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Consider all of the forces acting. In addition
to gravity and friction, there will be normal
forces Nf and Nw by the floor and wall
respectively on the ladder.
First sketch the FBD
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y
x
axis
L
sin φ mg + ( L − d ) sin φ Mg + L sin α F − L sin φ N f = 0
2
Nw
L/2
φ
x:
y:
Nw = F
Nf = Mg + mg
l
mg
m 
d
F = Mg tan φ  +

 L 2M 
Mg
F
Nf
Physics 201: Lecture 23, Pg 57
l
m 
d
 L + 2 M 
F = Mg tan φ
For a given coefficient of static friction µs,
the maximum force of friction F that can be
provided is µsNf = µs g(M + m).
v The ladder will slip if F exceeds
this value.
Cautionary note:
(1) Brace the bottom of ladders!
(2) Don’t make φ too big!
α
Nw
y
mg
F
Mg
α
Nf
Physics 201: Lecture 23, Pg 58
Example: Ladder against smooth wall
We have just calculated that
m
Substituting: Nf = Mg + mg and
solve for F :
d
d
φ
L/2
cosφ
Again use the fact that FNET = 0
in both x and y directions:
l
Since we are not interested in Nw, calculate
torques about an axis through the top end of
the ladder, in the z direction.
torque
φ
m
d
F
Physics 201: Lecture 23, Pg 59
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