Download Angular Momentum Solutions

Version 001 – Angular Momentum – smith – (3102F16B1)
This print-out should have 47 questions.
Multiple-choice questions may continue on
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before answering.
Airplane Momentum
001 (part 1 of 2) 10.0 points
An airplane of mass 18691 kg flies level to the
ground at an altitude of 14 km with a constant
speed of 178 m/s relative to the Earth.
What is the magnitude of the airplane’s
angular momentum relative to a ground observer directly below the airplane in kg ·m2/s?
Correct answer: 4.6578 × 1010 kg · m2 /s.
Explanation:
Since the observer is directly below the airplane,
L = hmv
002 (part 2 of 2) 10.0 points
Does this value change as the airplane continues its motion along a straight line?
1. Yes. L decreases as the airplane moves.
2. Yes. L increases as the airplane moves.
1
axis perpendicular to the circle and through
its center?
2
~ = 24 kg · m correct
1. kLk
s
2
~ = 12 m
2. kLk
s
~ =9 N·m
3. kLk
kg
2
~ = 13.5 kg · m
4. kLk
s2
~ = 18 N · m
5. kLk
kg
Explanation:
The angular momentum is
~ = m v r = (2 kg) (3 m/s) (4 m)
kLk
= 24 kg · m2 /s .
Conical Pendulum 04
004 10.0 points
A small metallic bob is suspended from the
ceiling by a thread of negligible mass. The
ball is then set in motion in a horizontal circle
so that the thread describes a cone.
3. Yes. L changes in a random pattern as
the airplane moves.
3. 2
m
5. No. L = constant. correct
Explanation:
L = constant since the perpendicular distance from the line of flight to Earth’s surface
doesn’t change.
AP B 1998 MC 6
003 10.0 points
A 2 kg object moves in a circle of radius 4 m
at a constant speed of 3 m/s. A net force of
4.5 N acts on the object.
What is the magnitude of the angular mo~ of the object with respect to an
mentum kLk
◦
9.8 m/s2
29
4. Yes. L changes with certain period as the
airplane moves.
v
9 kg
Calculate the magnitude of the angular momentum of the bob about a vertical axis
through the supporting point. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 40.5335 kg · m2 /s.
Explanation:
Version 001 – Angular Momentum – smith – (3102F16B1)
Let :
ℓ = 3.2 m ,
θ = 29◦ ,
g = 9.8 m/s2 ,
m = 9 kg .
and
Consider the free body diagram.
2
Decelerated Grinding Wheel
005 (part 1 of 2) 10.0 points
The motor driving a grinding wheel with a
rotational inertia of 0.8 kg m2 is switched
off when the wheel has a rotational speed
of 39 rad/s. After 6.7 s, the wheel has slowed
down to 31.2 rad/s.
What is the absolute value of the constant
torque exerted by friction to slow the wheel
down?
Correct answer: 0.931343 N m.
θ
Explanation:
We have
T
τ ∆t = ∆L = ∆(I ω) ,
so that
mg
Newton’s second law in the vertical and
horizontal projections, respectively, gives
T cos θ − m g
T cos θ
2
T sin θ − m ω ℓ sin θ
T
=0
= m g and
=0
= m ω2 ℓ ,
where the radius of the orbit is R = ℓ sin θ.
Dividing,
mg
m ω2 ℓ
r
g
ω=
.
ℓ cos θ
cos θ =
~r and ~v are perpendicular, where r =
ℓ sin θ , so the angular momentum L = m ~r ×~v
will be
L = m ω (ℓ sin θ)2
r
g ℓ3
= m sin2 θ
cos θ
2
= (9 kg) sin 29◦
r
(9.8 m/s2 ) (3.2 m)3
×
cos 29◦
= 40.5335 kg · m2 /s .
I |ω1 − ω0 |
∆t1
(0.8 kg m2 ) (39 rad/s − 31.2 rad/s)
=
6.7 s
= 0.931343 N m .
|τ | =
006 (part 2 of 2) 10.0 points
If this torque remains constant, how long after
the motor is switched off will the wheel come
to rest?
Correct answer: 33.5 s.
Explanation:
When the wheel comes to rest, its angular
speed is ω2 = 0; hence
I (ω0 − ω2 ) ∆t2 = τ
I ω0
=
|τ |
(0.8 kg m2 ) (39 rad/s)
=
(0.931343 N m)
= 33.5 s .
Holt SF 08Rev 66
007 (part 1 of 2) 10.0 points
a) Calculate the angular momentum of Earth
Version 001 – Angular Momentum – smith – (3102F16B1)
that arises from its spinning
motion on its
2
axis IE = 0.331ME RE .
Correct answer: 5.84082 × 1033 kg · m2 /s.
Explanation:
Basic Concept:
Given:
L = Iω = 0.331M R2 ω
R = 6.37 × 106 m
M = 5.98 × 1024 kg
ω = 1 rev/day
Merry Go Round and a Boy
009 10.0 points
The sketch shows the top view of a merrygo-round which is rotating clockwise. A boy
jumps on the merry-go-round in three different ways: I) from the left, II) from the top,
and III) from the right. For all three cases,
the boy lands on the same spot.
II
III
I
ω
Solution:
L = 0.331(5.98 × 1024 kg)
1 rev
6
2
· (6.37 × 10 m) ·
1 day
2π rad
1 day
1h
·
1 rev
24 h
3600 s
33
2
= 5.84082 × 10 kg · m /s
008 (part 2 of 2) 10.0 points
b) Calculate the average angular momentum
of Earth that arises from its orbital motion
about the sun.
Correct answer: 2.66465 × 1040 kg · m2 /s.
Explanation:
Basic Concept:
Given:
3
L = Iω = M r 2 ω
r = 1.496 × 1011 m
ω = 1 rev/365.25 days
Solution:
L = (5.98 × 1024 kg)(1.496 × 1011 m)2
1 rev
2π rad
·
365.25 day
1 rev
1h
1 day
·
24 h
3600 s
40
= 2.66465 × 10 kg · m2 /s
Compare the final angular momenta of the
merry-go-round for the three cases.
1. LIII > LII > LI
2. LI = LII = LIII
3. LI = LIII > LII
4. LI > LII > LIII correct
5. LII > LI = LIII
Explanation:
With respect to the center of the merrygo-round, the angular momentum of the boy
is: I) clockwise, II) 0, and III) counterclockwise. The angular momentum of the merrygo-round is along the clockwise direction. By
adding the two angular momenta in each case,
we get LI > LII > LIII .
Net L of point masses
010 10.0 points
Two objects are moving in the x-y plane as
shown.
Version 001 – Angular Momentum – smith – (3102F16B1)
The magnitude of their total angular momentum (about the origin O) is
counter-clockwise as the positive angular direction.
1. 8 kg · m2 /s
~ = m v0 R cos θ k̂
1. L
2. 24 kg · m2 /s
~ = m v0 R sin θ k̂
2. L
3. 32 kg · m2 /s
~ = m v0 R k̂
3. L
4. 0 kg · m2 /s correct
~ = m v0
4. L
5. 96 kg · m2 /s
~ = 0 correct
5. L
Explanation:
The angular momentum of a particle is
~ = I~ω , where we remember that
given by L
the ω is positive for counterclockwise rotation
in the x − y plane and negative for clockwise
rotation.
We can see above that the two particles
are rotating in opposite senses. For a point
particle, the moment of inertia is given by
I = mr 2 .
Consequently, we have:
4
R
k̂
2
Explanation:
~ = ~r × ~p = 0
L
since ~r = 0 .
012 (part 2 of 4) 10.0 points
Using the origin as the pivot, find the angular
momentum when the particle is at the highest
point of the trajectory.
m v0 2
sin θ cos2 θ
Lnet = L1 + L2
2g
= I 1 ω1 − I 2 ω2
m v0 3
~
2.
L
=
+
sin2 θ cos θ
= (m1 r12 )ω1 − (m2 r22 )ω2
2g
2
= (1 kg)(2 m)2 (4 rad/s) − (4 kg)(1 m)2 (4 rad/s)
~ = − m v0 sin2 θ cos θ
3.
L
2g
= 0 kg · m2 /s
2
~ = + m v0 sin2 θ cos θ
4. L
2g
Projectile Analysis 01
3
011 (part 1 of 4) 10.0 points
~ = − m v0 sin θ cos2 θ
5.
L
A particle of mass m moving in the gravi2g
tational field of the Earth is launched with
m v0 2
~
6. L = −
sin θ cos2 θ
an initial velocity vo at an angle θ with the
2g
horizontal.
3
~ = + m v0 sin θ cos2 θ
7. L
C
y
2g
vh
3
~ = − m v0 sin2 θ cos θ
8. L
2g
h
~ is in the ̂ direction.
θ
9. L
A
x
O
R
~ is in the ı̂ direction.
10. L
k̂
k̂
k̂
k̂
k̂
k̂
k̂
k̂ correct
vo
~ =+
1. L
vR
Using the origin as the pivot, find the angular momentum when it is at the origin. Use
Explanation:
Denote the position vector from the origin
to the highest point of the trajectory as ~rh .
Version 001 – Angular Momentum – smith – (3102F16B1)
At the highest point ~vh = ~vx = vo cos θ ı̂ and
~ = ~rh × ~ph = ~rh × m~vh
L
so that
~ = m |~vh | |~rh| sin θh .
|L|
The maximum height is
h = |~rh | sin θh =
vo2 sin2 θ
2g
and since |~vh | = |~v0x | = vx , the magnitude of
the angular momentum is
~ = m h vx = m
|L|
=
vo2 sin2 θ
(vo cos θ)
2g
m vo3
sin2 θ cos θ .
2g
The direction is −k̂ by the right hand rule of
the cross product.
013 (part 3 of 4) 10.0 points
Using the origin as the pivot, find the angular
momentum of the particle just before it hits
the ground.
2 m v0 3
sin2 θ cos θ
g
2
~ = − 2 m v0 sin θ cos2 θ
2. L
g
2 m v0 2
~
3. L = −
sin2 θ cos θ
g
3
~ = − 2 m v0 sin θ cos2 θ
4. L
g
~ is in the ı̂ direction.
5. L
~ =−
1. L
k̂ correct
Explanation:
Consider the motion to reach the maximum
height:
vf = 0 = vo sin θ − g t1
vo sin θ
t1 =
,
g
so the range of the projectile is
R = (vo cos θ) 2 t1 =
2 vo2 sin θ0 cos θ0
,
g
and the angular momentum at that position
is
2 m vo3
L = m vo sin θ R =
sin2 θ cos θ .
g
Using the right hand rule, the direction
should be −k̂ .
014 (part 4 of 4) 10.0 points
What torque causes its angular momentum to
change?
1. The torque from the initial force required
to shoot the particle.
k̂
k̂
k̂
2
~ = + 2 m v0 sin2 θ cos θ k̂
6. L
g
2
~ = + 2 m v0 sin θ cos2 θ k̂
7. L
g
3
~ = + 2 m v0 sin2 θ cos θ k̂
8. L
g
~ is in the ̂ direction.
9. L
~ =+
10. L
5
2 m v0 3
sin θ cos2 θ k̂
g
2. The torque from the centripetal acceleration.
3. The torque due to the initial velocity.
4. The torque due to the particle’s impulse.
5. The torque due to gravity. correct
Explanation:
Gravity is the only force acting on the particle. The change in angular momentum is
negative (going from zero to negative values)
because the torque of the gravity force is negative (−k̂ direction) as you can see from the
~ ×F
~.
right hand rule for cross product: ~τ = R
Time Dependent Momentum
015 10.0 points
Version 001 – Angular Momentum – smith – (3102F16B1)
The position vector of a particle of mass 2 kg
is given as a function of time by
~r = (1 m) ı̂ + (5 m/s) t ̂ .
Determine the magnitude of the angular
momentum of the particle with respect to the
origin at time 3 s .
Let : m = 5 kg ,
v = 2.5 m/s ,
r = 3 m.
and
The angular momemntum is
= 37.5 kg · m2 /s .
Explanation:
and
Basic Concepts:
~ = ~r × ~p
L
Solution:
∂ ~r
~v =
= vy
∂t
= (5 m/s) ̂ .
~ = ~r × ~p
L
= m ~r × ~v
∂ ~r
= m ~r ×
∂t
= m (~x ı̂ + ~y ̂) × ~v ̂
= m x vy k̂
= (2 kg) (1 m) (5 m/s) k̂
= 10 kg m2 /s k̂ ,
and is constant as a function of time since
̂ × ̂ = 0 .
Tipler PSE5 10 44
016 (part 1 of 3) 10.0 points
A 5 kg particle moves at a constant speed of
2.5 m/s around a circle of radius 3 m.
What is its angular momentum about the
center of the circle?
Correct answer: 37.5 kg · m2 /s.
Explanation:
L = m v r = (5 kg) (2.5 m/s) (3 m)
Correct answer: 10 kg m2 /s.
Let : ~r = x ı̂ + y ̂ ,
x = 1 m,
y = vy t = (5 m/s) t ,
t = 3 s.
6
017 (part 2 of 3) 10.0 points
What is its moment of inertia about an axis
through the center of the circle and perpendicular to the plane of the motion?
Correct answer: 45 kg · m2 .
Explanation:
The moment of inertia is
I = m r 2 = (5 kg)(3 m)2
= 45 kg · m2 .
018 (part 3 of 3) 10.0 points
What is the angular speed of the particle?
Correct answer: 0.833333 rad/s.
Explanation:
The angular speed is
L
37.5 kg · m2 /s
ω= =
I
45 kg · m2
= 0.833333 rad/s .
Tipler PSE5 10 79 02
019 (part 1 of 2) 10.0 points
A particle of mass 5 kg moves with velocity
~v = (2 m/s)ı̂ along the line z = 0 m, y = 1 m.
Find the angular momentum relative to the
origin when the particle is at x = 16 m, y =
1 m.
1. −(1.0 kg · m2 /s) ĵ
2. −(1.0 kg · m2 /s) k̂
Version 001 – Angular Momentum – smith – (3102F16B1)
3. −(2.0 kg · m2 /s) k̂
4. −(2.0 kg · m2 /s) ĵ
5. −(10 kg · m2 /s)ĵ
6. (47.7 kg · m2 /s) k̂
7. (47.7 kg · m2 /s) ĵ
8. −(10 kg · m2 /s)k̂ correct
Explanation:
Let : m = 5 kg ,
~v = 2 m/s , and
~r = (16 m)ı̂ + (1 m) ̂ .
The momentum is
~p = m~v = (5 kg) (2 m/s)ı̂ = (10 kg · m/s)ı̂ .
The angular momentum is
~ = ~r × ~p
L
~ = [(16 m)ı̂ + (1 m) ̂] × (10 kg · m/s)ı̂
L
= (10 kg · m2 /s) (̂ × ı̂)
= −(10 kg · m2 /s)k̂ .
020 (part 2 of 2) 10.0 points
~ = (−8 N)ı̂ is applied to the particle.
A force F
Find the torque relative to the origin due
to this force.
1. (20.0 N · m) ĵ
2. (−8 N · m) k̂
3. (10.0 N · m) ĵ
4. (8 N · m) k̂ correct
5. (−8 N · m) ĵ
6. (8 N · m) ĵ
7. (10.0 N · m) k̂
7
8. (20.0 N · m) k̂
Explanation:
~ = (−8 N)ı̂ .
Let : F
The torque due to the force is
~
~τ = ~r × F
= [(16 m)ı̂ + (1 m) ̂] × (−8 N)ı̂
= −(8 N · m) (̂ × ı̂)
= (8 N · m)k̂ .
AP M 1998 MC 32 33
021 (part 1 of 2) 10.0 points
A wheel with rotational inertia I is mounted
on a fixed, frictionless axle. The angular
speed ω of the wheel is increased from zero to
ωf in a time interval T .
What is the average net torque τ on the
wheel during this time interval?
I ωf
correct
T
I ωf
2. τ = 2
T
ωf
3. τ = 2
T
I ωf2
4. τ =
T
ωf
5. τ =
T
Explanation:
The change of angular momentum of the
wheel is
∆L = I ωf ,
1. τ =
so the average net torque on the wheel during
this time interval is
τ=
I ωf
L
=
.
T
T
022 (part 2 of 2) 10.0 points
What is the average power input to the wheel
during this time interval?
Version 001 – Angular Momentum – smith – (3102F16B1)
I 2 ωf
1. P =
2T2
I ωf2
correct
2. P =
2T
I ωf
3. P =
2T
I ωf2
4. P =
2T2
I 2 ωf2
5. P =
2T2
Explanation:
The change in the energy of the wheel is
1
I ωf2 in the time interval T , so the average
2
power input to the wheel is
I ωf2
1
2 1
I ωf =
.
2
T
2T
Rotating Meter Stick
023 (part 1 of 2) 10.0 points
A particle of mass 0.31 kg is attached to
the 100-cm mark of a meter stick of mass
0.35 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of
1.1 rad/s.
Calculate the angular momentum of the
system when the stick is pivoted about an
axis perpendicular to the table through the
50-cm mark.
8
and the angular momentum is
L1 = I1 ω
= (0.106667 kg m2 )(1.1 rad/s)
= 0.117333 kg m2 /s .
024 (part 2 of 2) 10.0 points
Calculate the angular momentum of the system when the stick is pivoted about an axis
perpendicular to the table through the 0-cm
mark.
Correct answer: 0.469333 kg m2 /s.
Explanation:
The moment of inertia of the stick in this
1
case is Is = ms (1 m)2 and of the particle
3
attached to the stick Ip = mp (1 m)2 . The
total moment of inertia of the system is
I2 = Is + Ip
1
= (0.35 kg) (1 m)2 + (0.31 kg (1 m)2
3
= 0.426667 kg m2 ,
and the angular momentum is
L2 = I2 ω
= (0.426667 kg m2 )(1.1 rad/s)
= 0.469333 kg m2 /s .
Correct answer: 0.117333 kg m2 /s.
Explanation:
L = Iω
The moment of inertia of the stick is
1
ms (1 m)2 and of the particle atIs =
12
2
1
m . The
tached to the stick Ip = mp
2
total moment of inertia of the system is
I1 = Is + Ip
1
=
(0.35 kg) (1 m)2 + (0.31 kg)
12
= 0.106667 kg m2 ,
1
m
2
2
Rotor of a Generator
025 10.0 points
A net torque of magnitude 200 N m is exerted
on the rotor of an electric generator for 32 s.
What is the magnitude of the angular momentum of the rotor at the end of 32 s if it
was initially at rest?
Correct answer: 6400 kg m2 /s.
Explanation:
We have:
∆L = τnet ∆t
Version 001 – Angular Momentum – smith – (3102F16B1)
Since the rotor was initially at rest, Li = 0,
Lf = (200 N m)(32 s)
9
027 (part 2 of 2) 10.0 points
What is the total angular momentum of the
system about the z-axis relative to point B
along y axis if d2 = 25 m?
= 6400 kg m2 /s
Total Angular Momentum
026 (part 1 of 2) 10.0 points
Two particles move in opposite directions
along a straight line. Particle 1 of mass
m1 = 22 kg at x1 = 16 m moves with a
speed v1 = 48 m/s (to the right), while the
particle 2 of mass m2 = 73 kg at x2 = −24 m
moves with a speed v2 = −26 m/s (to the
left).
Given: Counter-clockwise is the positive
angular direction.
Correct answer: −21050 kg m2 /s.
Explanation:
Since at point B
r sin θ = d2 ,
we find
B
l B = d2 p
= d2 (m1 v1 + m2 v2 )
= (25 m) [(22 kg)(48 m/s)
+ (73 kg) (−26 m/s)]
= −21050 kg m2 /s .
d2
v2
v1
m2
m1
x2
x1
d1
A
y
x
What is the total angular momentum of
the system about the z-axis relative to point
A along y axis if d1 = 13 m?
Correct answer: 10946 kg m2 /s.
Explanation:
The angular momentum along z axis is
given by
L = r p sin θ ,
where p is the total linear momentum. We
find that
p = m1 v1 + m2 v2 ,
where v2 is negative. Since at point A
r sin θ = −d1 ,
we find
l A = d1 p
= −d1 (m1 v1 + m2 v2 )
= (−13 m) [(22 kg)(48 m/s)
+ (73 kg) (−26 m/s)]
= 10946 kg m2 /s .
Bullet Rotates a Rod 01
028 (part 1 of 2) 10.0 points
A wooden block of mass M hangs from a rigid
rod of length ℓ having negligible mass. The
rod is pivoted at its upper end. A bullet of
mass m traveling horizontally and normal to
the rod with speed v hits the block and gets
embedded in it.
ℓ
v
m
M
What is the angular momentum L of the
block-bullet system, with respect to the pivot
point immediately after the collision?
1. L = m v ℓ correct
Mm
vℓ
2. L =
M +m
Version 001 – Angular Momentum – smith – (3102F16B1)
3. L = (M − m) v ℓ
4. L = (m + M ) v ℓ
5. L = M v ℓ
Explanation:
X
X
~ = 0.
If
~τext = 0, then
L
The net angular momentum of the system
conserves, and
Li = Lf = L = m v ℓ .
029 (part 2 of 2) 10.0 points
Kf
What is the fraction
(the final kinetic
Ki
energy compared to the initial kinetic energy)
in the collision?
Kf
2m
=
Ki
m+M
Kf
M
=
2.
Ki
M +m
Kf
M
=
3.
Ki
M −m
Kf
m
=
correct
4.
Ki
m+M
Kf
m
5.
=
Ki
M −m
Explanation:
By conservation of the angular momentum
1.
Li = Lf = L
m v ℓ = (m + M ) vf ℓ
m
vf = v
m+M
1
1
m v 2 and Kf = I ωf2 where
2
2
vf
2
, so
I = (M + m) ℓ and ωf =
ℓ
1
Kf = (M + m) vf2 and
2
Ki =
1 m2
v2
Kf
m
2 M +m
=
=
.
1
Ki
M +m
2
mv
2
10
Child on a Merrygoround
030 10.0 points
A child is standing on the edge of a merry-goround that is rotating with frequency f. The
child then walks towards the center of the
merry-go-round.
For the system consisting of the child plus
the merry-go-round, what remains constant
as the child walks towards the center? (neglect friction in the bearing)
1. neither mechanical energy nor angular
momentum
2. only mechanical energy
3. mechanical energy and angular momentum
4. only angular momentum correct
Explanation:
The forces external to the system are not
exerting any torque, so the angular momentum is conserved. On the other hand, the friction force acting on the child is doing work, because she is moving towards the center (which
is the direction of that force).
Child on a MerryGoRound 02 v1
031 10.0 points
A child of mass 52.8 kg sits on the edge of
a merry-go-round with radius 1.7 m and moment of inertia 134.281 kg m2 . The merrygo-round rotates with an angular velocity of
1.9 rad/s. The child then walks towards the
center of the merry-go-round and stops at a
distance 0.612 m from the center. Now what
is the angular velocity of the merry-go-round?
Correct answer: 3.53803 rad/s.
Explanation:
When the child moves inward, the moment of
inertia of the system iM GR + ichild (the merrygo-round plus the child) changes. Therefore,
to conserve angular momentum, the angular
Version 001 – Angular Momentum – smith – (3102F16B1)
velocity of the system must change. Specifically:
Linit = Lf inal
(ichild + iM GR ) ω = (ichild,2 + iM GR ) ω2
The moment of inertia of the child is m r 2 .
Therefore
m r 2 + iM GR
ω2 =
ω.
m r2 2 + iM GR
Child on Merry Go Round
032 10.0 points
A playground merry-go-round has a radius of
3 m and a rotational inertia of 600 kg · m2 /s.
It is initially spinning at 0.8 rad/s when a
20 kg child crawls from the center to the rim.
When the child reaches the rim the angular
velocity of the merry-go-round is:
033 10.0 points
A thin uniform cylindrical turntable of radius 1.8 m and mass 23 kg rotates in a horizontal plane with an initial angular speed of
8.5 rad/s. The turntable bearing is frictionless. A clump of clay of mass 5.2 kg is dropped
onto the turntable and sticks at a point 0.96 m
from the point of rotation.
Find the angular speed of the clay and
turntable.
Correct answer: 7.53133 rad/s.
Explanation:
Let : R = 1.8 m ,
x = 0.96 m ,
M = 23 kg ,
ω = 8.5 rad/s ,
m = 5.2 kg .
1. 0.89 rad/s
3. 0.8 rad/s
X
4. 0.62 rad/s correct
Explanation:
We conserve angular momentum, realizing
that the child’s moment of inertia is given by
mr 2 :
and
Basic Concepts:
2. 1.1 rad/s
5. 0.73 rad/s
11
~ = d ~τext
L
dt
∆K = Kf − Ki = Q .
From conservation of the angular momentum
it follows that
Li = Lf
or
I i ω = I f ωf ,
Lf = L0
where
I f ωf = I 0 ω0
1
2
Ii = M R 2
I + mrf ωf = I + mr02 ω0
2
1
I
= (23 kg) (1.8 m)2
ωf =
ω0
2
2
I + mr
2
=
37.26
kg m2 .
600 kg · m /s
=
600 kg · m2 /s + (20 kg)(3 m)2
Similarly,
× (0.8 rad/s)
= 0.62 rad/s
I f = I i + m x2
Clay on a Turntable 01
= (37.26 kg m2 ) + (5.2 kg) (0.96 m)2
= 42.0523 kg m2 .
Version 001 – Angular Momentum – smith – (3102F16B1)
12
Finally,
5. II and V only
Ii
ωf =
ω
If
(37.26 kg m2 )
=
(8.5 rad/s)
(42.0523 kg m2 )
= 7.53133 rad/s .
6. II, III and IV only
7. III, IV and V only
8. I, II and III only
9. II, III, IV and V only
Clay Rotates a Rod 01a
034 10.0 points
A uniform rod has a mass 2 m and a length
ℓ, and it can spin freely in a horizontal plane
about a pivot point O at the center of the rod.
A piece of clay with mass m and velocity v
hits one end of the rod, and causes the rodclay system to spin. Viewed from above the
scheme is as follows:
v
m
ℓ
0
ω
ω
2m
(a) before
(b) during
(c) after
After the collisions the rod and clay system
has an angular velocity ω about the pivot.
Which quantity/quantities
I) total mechanical energy
II) total linear momentum
III) total angular momentum with respect to
pivot point O
IV) total gravitational potential energy
V) total kinetic energy
is/are conserved in this process?
1. All of these
2. None of these
3. III and IV only correct
4. II and III only
10. I only
Explanation:
V) Total KE is not conserved. One can
Kf
Ii
verify that
=
< 1. The inequality
Ki
If
here is due to the fact that Ii is the moment
of inertia of the clay and If is the moment of
inertia of the clay+rod.
IV) Since the motion is in a horizontal plane
there is no change in the potential energy.
IV) and V) together imply that the total
mechanical energy (KE + PE) is not conserved.
II) Linear momentum is not conserved by
inspection. Apparently the linear momentum is altered by the presence of the pivot.
The pivot provides an external force to the
rod+clay system.
III) Angular momentum is conserved, since
there is no external torque applied to the
rod+clay system.
Clay Rotates a Rod 01
035 (part 1 of 2) 10.0 points
A uniform rod, supported and pivoted at
its midpoint, but initially at rest, has a mass
2 m and a length ℓ. A piece of clay with mass
m and velocity v0 hits one end of the rod, gets
stuck and causes the clay-rod system to spin
about the pivot point O at the center of the
rod in a horizontal plane. Viewed from above
the scheme is
Version 001 – Angular Momentum – smith – (3102F16B1)
plus the moment of inertia of the clay:
2
1
ℓ
2
If = Irod + Iclay =
2mℓ + m
12
2
5
m ℓ2 .
=
12
v0
m
ℓ
0
ω
ω
(a) 2 m
Before
(b)
During
(c)
After
With respect to the pivot point O, what
is the magnitude of the initial angular momentum Li of the piece of clay and the final
moment of inertia If of the clay-rod system?
After the collisions the clay-rod system has an
angular velocity ω about the pivot.
1. Li = m v
ℓ
,
2
2. Li = m v ℓ ,
3. Li = m v ℓ ,
4. Li = m v ℓ ,
ℓ
,
2
ℓ
6. Li = m v ,
2
ℓ
7. Li = m v ,
2
5. Li = m v
8. Li = m v ℓ ,
13
7
m ℓ2
12
3
If =
m ℓ2
12
5
If =
m ℓ2
12
8
m ℓ2
If =
12
4
If =
m ℓ2
12
5
If =
m ℓ2 correct
12
8
If =
m ℓ2
12
4
If =
m ℓ2
12
If =
Explanation:
Since the total external torque acting on
the clay-rod system is zero, the total angular
momentum is a constant of motion. The total
initial angular momentum Li is simply the
angular momentum of the clay, since the rod
is at rest initially
mvℓ
.
Li = |~r × ~p| = m r v =
2
The final moment of inertia If of the clayrod system is the moment of inertia of the rod
036 (part 2 of 2) 10.0 points
What is the final angular speed ωf of the
rod-clay system¿
3 v
5 ℓ
6
2. ωf = v
5
5 v
3. ωf =
12 ℓ
4
4. ωf = v
6
12 v
5. ωf =
7 ℓ
5 v
6. ωf =
6 ℓ
12
v
7. ωf =
7
6 v
8. ωf =
correct
5 ℓ
6 v
9. ωf =
2 ℓ
12 v
10. ωf =
5 ℓ
Explanation:
The total angular momenta are the same:
1. ωf =
Lf = Li
mvℓ
I f ωf =
2
m
vℓ
5
m ℓ 2 ωf =
12
2
6 v
ωf =
.
5 ℓ
Concept 08 43
037 10.0 points
You sit at the middle of a large turntable at
an amusement park as it is set spinning
on nearly frictionless bearings, and then
allowed to spin freely.
Version 001 – Angular Momentum – smith – (3102F16B1)
When you crawl toward the edge of the
turntable, does the rate of the rotation increase, decrease, or remain unchanged, and
why?
1. increases; conservation of momentum
2. decreases; conservation of energy
3. increases; conservation of energy
4. decreases; conservation of momentum
correct
Explanation:
The rotational inertia of the system (you
and the rotating turntable) is least when you
are at the rotational axis. As you move outward, the rotational inertia of the system increases.
Applying conservation of angular momentum, as you move toward the outer rim, you
increase the rotational inertia of the spinning
system, thus decreasing the angular speed.
Also, if you don’t slip as you move out, you
exert a friction force on the turntable opposite to its direction of rotation, thereby also
slowing it down.
Concept 08 45
038 10.0 points
Strictly speaking, as more and more skyscrapers are built on the surface of the Earth, does
the day tend to become longer or shorter?
And strictly speaking, does the falling of autumn leaves tend to lengthen or shorten the
24-hour day? What physical principle supports your answers?
1. lengthen; shorten; conservation of angular
momentum correct
2. shorten; lengthen; conservation of inertia
3. lengthen; shorten; conservation of kinetic
energy
4. shorten; lengthen; conservation of angular
torque
14
Explanation:
Applying conservation of angular momentum, as the radial distance of mass increases,
the angular speed decreases. The mass of material used to construct skyscrapers is lifted,
slightly increasing the radial distance from the
Earth’s spin axis. This would tend to slightly
decrease the earth’s rate of rotation, which
in turn tends to lengthen days slightly. The
opposite effect occurs for falling leaves, since
their radial distance from the earth’s axis decreases. As a practical matter, these effects
are entirely negligible!
Conceptual 07 05
039 10.0 points
Why does a helicopter have a tail rotor?
1. This keeps the helicopter from spinning
out of control. correct
2. None of these
3. This makes the helicopter move faster.
4. The tail rotor is just for decoration.
5. This increases safety if any of the rotors
breaks down.
Explanation:
Without the tail rotor, the body would spin
in the opposite direction of the rotors.
Conceptual Rotating KE
040 10.0 points
How can the rotational kinetic energy Kr of
a rigid body be written in terms of the magnitude L fo its angular momentum L and its
rotational moment of inertia I?
L2
correct
2I
1
2. Kr = I L
2
√
3. Kr = 2 L I
1. Kr =
4. Kr =
1
I L2
2
Version 001 – Angular Momentum – smith – (3102F16B1)
L
I
Explanation:
For a symmetric rigid body, we know that
~ = I ~ω . Putting this relation into the definiL
tion of angular rotational kinetic energy,
5. Kr =
1
I ω2
2
2
L
1
= I
2
I
2
L
=
2I
15
So,
KEf
=
KEi
L2f
2 If
L2i
2 Ii
=
Ii
4
Ii
=
=
If
0.75 Ii
3
Kr =
Figure Skater
041 (part 1 of 2) 10.0 points
A figure skater rotating on one spot with both
arms and one leg extended has moment of
inertia Ii . She then pulls in her arms and the
extended leg, reducing her moment of inertia
to 0.75 Ii.
What is the ratio of her final to initial kinetic energy?
042 (part 2 of 2) 10.0 points
Consider the following statements for the figure skater:
I. Angular momentum was conserved.
II. Mechanical energy was conserved.
III. The kinetic energy changed because of
energy dissipation due to friction.
IV. Her rotation rate changed in response
to a torque exerted by pulling in her arms and
leg.
Which is the correct combination of statements?
1. I, II, IV
2. I and II
1. 4/3 correct
3. II
2. 1
4. I correct
3. 9/16
5. I, II, III
4. 3/8
5. 1/2
6. 16/9
7. 8/3
8. 3/4
9. 2
Explanation:
The angular momentum was conserved, so
Li = Lf
1
I ω2,
2
L2
KEr =
2I
Since L = I ω and KEr =
Explanation:
(I) The angular momentum was conserved
since there was no external torque acting on
the skater.
(II) From part 1, we found that the kinetic
energy increased. and the potential energy
didn’t change, so the total mechanical energy
increased. (II) is wrong.
(III) We found that the total kinetic energy
increased, and therefore, we can infer that
there was no energy dissipation.
(IV) The force pulling her arms in would be
perpendicular to her rotation, and therefore,
exert no torque.
Ergo, the only correct statement is (I).
Figure Skater Spin
Version 001 – Angular Momentum – smith – (3102F16B1)
043 (part 1 of 2) 10.0 points
A figure skater on ice spins on one foot. She
pulls in her arms and her rotational speed
increases.
Choose the best statement below:
1. Her angular speed increases because air
friction is reduced as her arms come in.
2. Her angular speed increases because her
angular momentum increases.
3. Her angular speed increases because her
potential energy increases as her arms come
in.
4. Her angular speed increases because her
angular momentum is the same but her moment of inertia decreases. correct
5. Her angular speed increases because by
pulling in her arms she creates a net torque in
the direction of rotation.
6. Her angular speed increases due to a net
torque exerted by her surroundings.
Explanation:
The initial angular momentum of the figure
skater is Ii ωi . After she pulls in her arms, the
angular momentum of her is If ωf . Note that
If < Ii because her arms now rotate closer to
the rotation axis and reduce the moment of
inertia. Since the net external torque is zero,
angular momentum remains unchanged, and
so Ii ωi = If ωf = L. Therefore, ωf > ωi .
044 (part 2 of 2) 10.0 points
And again, choose the best statement below:
1. When she pulls in her arms, her rotational potential energy increases as her arms
approach the center.
2. When she pulls in her arms, her moment
of inertia is conserved.
3. When she pulls in her arms, her rotational
kinetic energy must decrease because of the
decrease in her moment of inertia.
16
4. When she pulls in her arms, the work
she performs on them turns into increased
rotational kinetic energy. correct
5. When she pulls in her arms, her rotational
kinetic energy is conserved and therefore stays
the same.
6. When she pulls in her arms, her angular momentum decreases so as to conserve
energy.
Explanation:
The kinetic energy of the figure skater,
E=
1
1
I ω2 = L ω .
2
2
Since ω increases after she pulls in her arms
as mentioned above, the total kinetic energy
increases. This additional energy comes from
the figure skater, namely she has to perform
some work to achieve this.
Ice Skater pulls arms in
045 10.0 points
An ice skater with rotational inertia I0 is spinning with angular velocity ω0 . She pulls her
arms in, decreasing her rotational inertia to
I0 /3. Her angular velocity becomes:
1. ω0 /3
√
2. ω0 / 3
3. ω0
4. 3ω0 correct
√
5. 3ω0
Explanation:
Angular momentum is given by L = Iω,
and is conserved on ice, so:
I0 ω0 = Iω
1
=
I0 ω
3
ω = 3ω0
Version 001 – Angular Momentum – smith – (3102F16B1)
Mimic a Spinning Skater 01
046 (part 1 of 2) 10.0 points
A student sits on a rotating stool holding two
2 kg objects. When his arms are extended
horizontally, the objects are 1 m from the axis
of rotation, and he rotates with angular speed
of 0.61 rad/sec. The moment of inertia of the
student plus the stool is 6 kg m2 and is assumed to be constant. The student then pulls
the objects horizontally to a radius 0.27 m
from the rotation axis.
ωi
ωf
17
From conservation of the angular momentum
it follows that
I i ωi = I f ωf
Ii
ωf = ωi
If
= (0.61 rad/sec)
10 kg m2
6.2916 kg m2
= 0.969547 rad/s .
047 (part 2 of 2) 10.0 points
Calculate the change in kinetic energy of the
system.
Correct answer: 1.09662 J.
Explanation:
(a)
(b)
Calculate the final angular speed of the
student.
Correct answer: 0.969547 rad/s.
Explanation:
Let : m = 2 kg ,
R = 1 m,
r = 0.27 m ,
ω = 0.61 rad/sec ,
Is = 6 kg m2 .
X
= 1.09662 J .
and
~ = Constant.
L
1
I ω2
2
The initial moment of inertia of the system is
Krot =
Ii = Is + 2 m R 2
= (6 kg m2 ) + 2 (2 kg) (1 m)2
= 10 kg m2 .
The final moment of inertia of the system is
If = Is + 2 m r 2
= 6 kg m2 + 2 (2 kg) (0.27 m)2
= 6.2916 kg m2 .
∆K = Kf − Ki
1
1
= If ωf2 − Ii ωi2
2
2
1
6.2916 kg m2 (0.969547 rad/s)2
=
2
1
− 10 kg m2 (0.61 rad/sec)2
2
= (2.95712 J) − (1.8605 J)