Version PREVIEW – Practice 8 – carroll – (11108)
This print-out should have 12 questions.
Multiple-choice questions may continue on
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before answering.
Inertia of Solids 01
001 10.0 points
A circular disk, a ring, and a square have the
same mass M and width 2 r.
ring
1
002 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.
The acceleration of gravity is 9.8 m/s2 .
square
disk
◦
34
2. 4
9.8 m/s2
m
2r
2r
2r
For the moment of inertia about their center of mass about an axis perpendicular to the
plane of the paper, which statement concerning their moments of inertia is true?
1. Isquare > Iring > Idisk correct
2. Idisk > Isquare > Iring
v
5 kg
Calculate the magnitude of the angular
momentum of the bob about the supporting point.
Correct answer: 19.9865 kg m2 /s.
Explanation:
Basic Concepts: Angular Momentum
3. Idisk > Iring > Isquare
4. Isquare > Idisk > Iring
L = m ~r × ~v .
5. Iring > Idisk > Isquare
6. Iring > Isquare > Idisk
Explanation:
In the ring, the same mass of the disk is
concentrated at the maximum distance from
the axis, so
Iring > Idisk .
In the square, the same mass of the ring
√ lies
at distances which are between r and r 2 (at
least the radius of the ring), so
Isquare > Iring
and
Note: In this problem ~r and ~v are perpendicular, where r = ℓ sin θ .
Let : ℓ = 2.4 m ,
θ = 34◦ ,
g = 9.8 m/s2 ,
m = 5 kg .
and
Isquare > Iring > Idisk .
Conical Pendulum 04
Use the free body diagram below.
Version PREVIEW – Practice 8 – carroll – (11108)
θ
T
2
bar is fastened by a pivot at one end to a
wall which is at an angle θ with respect to
the horizontal. The bar is held horizontal by
a vertical cord that is fastened to the bar at
a distance a distance xcord from the wall. A
mass m2 is suspended from the free end of the
bar.
T
θ
mg
Solution: The second Newton’s law in the
vertical and horizontal projections, respectively, in our case reads
T cos θ − m g = 0
T sin θ − m ω 2 ℓ sin θ = 0 ,
where T is the force with which the wire acts
on the bob and the radius of the orbit is
R = ℓ sin θ. From this system of equations we
find
r
g
ω=
ℓ cos θ
s
(9.8 m/s2 )
=
(2.4 m) cos(34◦ )
= 2.21932 rad/s .
The angular momentum L then will amount
to
L = m ω (ℓ sin θ)2
r
g ℓ3
2
= m sin θ
cos θ
2
= (5 kg) sin (34◦ )
s
(9.8 m/s2 ) (2.4 m)3
×
cos(34◦ )
= 19.9865 kg m2 /s .
Beam on a Slanted Wall 01
003 (part 1 of 2) 10.0 points
A solid bar of length L has a mass m1 . The
m1
xcord
m2
L
Find the tension T in the cord.
1. T = 0
1
L
2. T =
m1 + m2
g correct
2
xcord
3. T = (m1 + m2 ) g cos θ
1
L
4. T =
m1 + m2
g cos θ
2
xcord
g
L
5. T = (m1 + m2 )
x
2
cord
L
1
g
6. T = m1 + m2
2
xcord
1
L
7. T = m1 + m2
g cos θ
2
xcord
L
1
g sin θ
8. T = m1 + m2
2
xcord
L
1
m1 + m2
g sin θ
9. T =
2
xcord
10. T = (m1 + m2 ) g sin θ
Explanation:
Basic Concepts: Static Equilibrium:
X
F =0.
X
τ =0.
Torque: τ = r⊥ F = r F⊥
Version PREVIEW – Practice 8 – carroll – (11108)
We can begin by either summing the
torques or the forces and setting them equal to
zero. In this case, since we know little about
the reaction force between the bar and the
wall, it is easiest to begin by examining the
torques around the connection between the
bar and the wall, because the reaction forces
will produce no torques around that point.
X
L
τ = −L m2 g + xcord T − m1 g = 0 .
2
Solving this for T gives
L
1
m1 + m2
T =
g .
2
xcord
004 (part 2 of 2) 10.0 points
Find the the horizontal component of the
force exerted on the bar by the wall. (Take
the positive direction to be right.)
1. Fx = (m1 + m2 ) g cos θ
2. Fx = (m1 + m2 ) g sin θ
1
L
3. Fx =
m1 + m2
g cos θ
2
xcord
g
L
4. Fx = (m1 + m2 )
x
2
cord
1
L
5. Fx = m1 + m2
g sin θ
2
xcord
L
1
g
6. Fx = m1 + m2
2
xcord
L
1
m1 + m2
g sin θ
7. Fx =
2
xcord
8. Fx = 0 correct
L
1
m1 + m2
9. Fx =
g
2
xcord
L
1
g cos θ
10. Fx = m1 + m2
2
xcord
Explanation:
In general, there will be a horizontal reaction force Rx at the connection between an
object and a wall. However, none of the other
forces in this situation act in the horizontal
direction. Therefore:
X
Fx = Rx = 0 .
3
Rotation of a Solid Disk 03
005 10.0 points
A uniform solid disk of radius 6.85 m and
mass 34.8 kg is free to rotate on a frictionless
pivot through a point on its rim.
Pivot
6.85 m
If the disk is released from rest in the
position shown by the solid circle, what is
the speed of its center of mass when the
disk reaches the position indicated by the
dashed circle? The acceleration of gravity
is 9.8 m/s2 .
Correct answer: 9.4608 m/s.
Explanation:
Let :
r = 6.85 m and
m = 34.8 kg .
From the parallel axis theorem
3
1
I = M R2 + M R2 = M R2 .
2
2
so the final kinetic energy is
1 3
2
ω2
Kf =
MR
2 2
1 3
M v2
=
2 2
3
3
= M v 2 = M R2 ω 2 .
4
4
From conservation of energy
(K + U )i = (K + U )f
1
0 + M g R = I ω2
2
3
M g R = M R2 ω 2
4
r
4g
ω=
3R
Version PREVIEW – Practice 8 – carroll – (11108)
Hence the velocity of the center of mass is
vcm = ω R
r
Rg
3
r
(6.85 m) (9.8 m/s2 )
=2
3
= 9.4608 m/s .
=2
Alternate Solution The kinetic energy is
Kftotal = Kfrotational + Kfcm linear
4
Explanation:
From conservation of energy we have
Ui = Ktrans,f + Krot,f
1
1
M g h = M v2 + I ω2
2
2 2
1 2
v
1
2
2
MR
= Mv +
2
2 5
R2
r
7
10
=
M v 2 .v1
=
gh
10
7
r
10
(9.8 m/s2 ) (1.6 m)
=
7
= 4.73286 m/s .
1
1 1
2
M R ω2 + M v2
Kf =
2 2
2
1 1
1
=
M v2 + M v2
2 2
2
3
3
= M v 2 = M R2 ω 2 .
4
4
007 (part 2 of 2) 10.0 points
Calculate the speed of the sphere if it reaches
the bottom of the incline by slipping frictionlessly without rolling.
Correct answer: 5.6 m/s.
Explanation:
From conservation of energy we have
keywords:
Solid Sphere on an Incline 04
006 (part 1 of 2) 10.0 points
A solid sphere of radius 34 cm is positioned
at the top of an incline that makes 26 ◦ angle
with the horizontal. This initial position of
the sphere is a vertical distance 1.6 m above
its position when at the bottom of the incline.
The sphere is released and moves down the
incline.
34 cm
M
µ
26 ◦
1.6 m
ℓ
Calculate the speed of the sphere when it
reaches the bottom of the incline if it rolls
without slipping. The acceleration of gravity
is 9.8 m/s2 . The moment of inertia of a sphere
with respect to an axis through its center is
2
M R2 .
5
Correct answer: 4.73286 m/s.
Ui = Ktrans,f
1
M g h = M v2
2
p
v2 = 2 g h
q
= 2 (9.8 m/s2 ) (1.6 m)
= 5.6 m/s .
keywords:
/* If you use any of these, fix
the comment symbols.
g=9.8 ; —*
u=m/s2 ∗ |h = 6.626075e − 34; | ∗ u = J s ∗
|Someof yourvariablesresembleconstants.
Airplane Momentum
008 (part 1 of 2) 10.0 points
An airplane of mass 24172 kg flies level to the
ground at an altitude of 16 km with a constant
speed of 172 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: 6.65213 × 1010 kg · m2 /s.
Version PREVIEW – Practice 8 – carroll – (11108)
Explanation:
Since the observer is directly below the airplane,
L = hmv
009 (part 2 of 2) 10.0 points
Does this value change as the airplane continues its motion along a straight line?
1. Yes. L changes with certain period as the
airplane moves.
2. Yes. L decreases as the airplane moves.
First, calculate the planet’s angular momentum (relative to its spin axis) before the
impact.
Correct answer: 1.25797 × 1033 kg m2 /s.
Explanation:
Basic Concept: A rigid body rotating
around a fixed axis has angular momentum
L=I ×ω
where I is the body’s moment of inertia about
the axis of rotation.
Approximating the planet as a solid ball of
uniform density, its moment of inertia is
3. Yes. L increases as the airplane moves.
5. Yes. L changes in a random pattern as
the airplane moves.
Explanation:
L = constant since the perpendicular distance from the line of flight to Earth’s surface
doesn’t change.
Asteroid Collision
010 (part 1 of 3) 10.0 points
Consider an Earth-like planet hit by an asteroid.
The planet has mass Mp = 6.72 × 1023 kg
and radius Rp = 6.55 × 106 m, and you may
approximate it as a solid ball of uniform
density. It rotates on its axis once every
T = 16 hr. The asteroid has mass Ma =
3.99 × 1017 kg and speed va = 22500 m/s
(relative to the planet’s center); its velocity
vector points θ = 70◦ below the Eastward horizontal. The impact happens at an equatorial
location. The picture below shows the view
from above the planet’s North pole:
vm
θ
R
2
M R2
5
= 1.15322 × 1037 kg m2 .
Ip =
4. No. L = constant. correct
ω
5
Its angular velocity before the impact is
2π
T
= 0.000109083 rad/s , so
L=I ×ω
=I ×ω
= (1.15322 × 1037 kg m2 )
×(0.000109083 rad/s)
ω=
= 1.25797 × 1033 kg m2 /s .
011 (part 2 of 3) 10.0 points
Calculate the asteroid’s angular momentum
relative to the planetary axis.
Correct answer: 2.01117 × 1028 kg m2 /s.
Explanation:
Approximating the asteroid as a pointlike
particle, its angular momentum is
~a = R
~ × Ma ~va
L
~ is the asteroid’s radius vector and
where R
Ma ~va is its linear momentum vector.
At the moment of impact, both the radius
~ and the momentum vector Ma~va
vector R
of the asteroid lie in the planet’s equatorial
plane. Consequently, their vector product
is perpendicular to the equatorial plane and
Version PREVIEW – Practice 8 – carroll – (11108)
parallel to the planet axis. Because the horizontal component of the asteroid’s velocity is
directed Eastward — the same as the planet’s
rotation — the asteroid’s angular momentum
~ a has the same direction as the planet’s anL
~ p.
gular momentum L
In magnitude,
La = Rp Ma × horizontal component of va
= Rp Ma × va cos θ
6
and since we assume unchanged moment of inertia Ip′ = Ip , it follows that after the impact,
the planet rotates at new angular velocity
L′p
ω =
Ip
′
=ω+
Rp Ma va cos θ
.
Ip
In other words,
6
= (6.55 × 10 m)
×(3.99 × 1017 kg) (22500 m/s) cos(70◦ )
= 2.01117 × 1028 kg m2 /s .
012 (part 3 of 3) 10.0 points
The impact is totally inelastic — the asteroid
is stuck in the planet’s crust. But thanks to
the asteroid’s angular momentum, the planet
rotates faster after the impact than it did
before.
By how many seconds has the collision
shortened the planetary day?
For simplicity, ignore the effect of the asteroid’s mass on the planet’s moment of inertia
and assume
before
I after = Iplanet
.
Warning: At intermediate stages of this calculation, do not round off intermediate results
and keep at least 7 significant digits.
Correct answer: 0.920862 s.
Explanation:
The net angular momentum is conserved in
this collision, so after the impact the planet
(with the asteroid stuck in its crust) has
~ ′p = L
~p +L
~a .
L
~
Focusing on the Northward component of L
(the other two components vanish), we have
L′p = Lp + La
= Ip ω + Rp Ma va cos θ .
At the same time,
L′p = IP′ ω ′ ,
∆ω = ω ′ − ω
Rp Ma va cos θ
=
Ip
5 Ma va cos θ
=
2 Mp Rp
= 1.74396 × 10−9 rad/s .
In terms of the planetary day, this change
means
−∆T = T − T ′
2π 2π
=
− ′
ω
ω
2 π(ω ′ − ω)
=
ω ω′
2 π ∆ω 1
=
ω + ∆ω ω
2 π (1.74396 × 10−9 rad/s)
=
(0.000109085 rad/s)
1
×
(0.000109083 rad/s)
= 0.920862 s ,
where
ω + ∆ω = (0.000109083 rad/s)
+(1.74396 × 10−9 rad/s)
= 0.000109085 rad/s .