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```xu (rx575) – HW08 – macdonald – (55475)
This print-out should have 26 questions.
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001 (part 1 of 2) 10.0 points
A record has an angular speed
19.9 rev/min.
What is its angular speed?
of
Let : Lh
Lm
Mh
Mm
1
= 2.79 m ,
= 4.54 m ,
= 58.4 kg ,
= 110 kg .
and
The hour hand makes 1 rev in 24 hours and
the minute hand in 60 minutes. The moment
of inertia of a thin rod about an axis through
one end is
M L2
I=
,
3
so the total rotational kinetic energy is
Explanation:
Let : ω = 19.9 rev/min .
2
Ih ωh2 Im ωm
+
2
2
2
mh Lh 2 mm L2m 2
=
ωh +
ωm
6
6
(58.4 kg) (2.79 m)2
=
6
2 2
1h
&times;
12 h
3600 s
(110 kg) (4.54 m)2
+
6
2 2
1 min
&times;
60 min
60 s
Ktot =
ω = (19.9 rev/min)
rev
1 min
60 s
002 (part 2 of 2) 10.0 points
Through what angle does it rotate in 1.55 s?
Explanation:
Let :
t = 1.55 s .
θ = ωt
003 10.0 points
The hour and minute hands of a tower clock
like Big Ben in London are 2.79 m and 4.54 m
long and have masses of 58.4 kg and 110 kg,
respectively.
Calculate the total rotational kinetic energy of the two hands about the axis of rotation. Model the hands as long thin rods.
Explanation:
= 0.00115269 J .
Note: The rotational kinetic energy of
the hour hand is negligible compared to the
minute hand.
keywords:
004 (part 1 of 2) 10.0 points
A racing car travels on a circular track of
radius 154 m, moving with a constant linear
speed of 42.3 m/s.
Find its angular speed.
Explanation:
Let :
v = 42.3 m/s
R = 154 m .
and
xu (rx575) – HW08 – macdonald – (55475)
The linear speed is
v = Rω
v
42.3 m/s
ω=
=
R
154 m
2
A dentist’s drill starts from rest. After 4.28 s
of constant angular acceleration it turns at a
rate of 19840 rev/min.
Find the drill’s angular acceleration.
005 (part 2 of 2) 10.0 points
Find the magnitude of its acceleration.
Explanation:
With the car moving at a constant speed,
there is no tangential acceleration, so the acceleration is purely radial:
ar =
v2
(42.3 m/s)2
=
= 11.6188 m/s2 .
R
154 m
006 10.0 points
Initially a wheel rotating about a ﬁxed axis at
a constant angular deceleration of 0.4 rad/s2
has an angular velocity of 0 rad/s and an
What is the angular position of the wheel
after 3.5 s?
Explanation:
Let :
Explanation:
Let : t = 4.28 s and
ωf = 19840 rev/min .
Since ω0 = 0,
ωf − ω0
ωf
α=
=
t
t
19840 rev/min 2 π 1 min
=
&middot;
&middot;
4.28 s
rev 60 s
008 (part 2 of 2) 10.0 points
Throughout what angle does the drill rotate
during this period?
Explanation:
θ = ω0 t +
1
α t2
2
1
2
=0+
θ0 = 8.3 rad , and
t = 3.5 s .
From kinematics,
1 2
α t + θ0
2
1
2
009 (part 1 of 4) 10.0 points
Consider a rod of length L and mass m which
is pivoted at one end. An object with mass m
is attached to the free end of the rod.
θ = ω0 t +
007 (part 1 of 2) 10.0 points
m
C
L
20◦
m
xu (rx575) – HW08 – macdonald – (55475)
Determine the moment of inertia of the
system with respect to the pivot point. The
acceleration of gravity g = 9.8 m/s2 . Consider
the mass at the end of the rod to be a point
particle.
3
m L2
2
4
2. I = m L2 correct
3
1. I =
3. None of these.
5
m L2
3
13
m L2
5. I =
12
4. I =
6. I = L2
5
m L2
4
Explanation:
The moment of inertia of the rod with respect to the pivot point is
7. I =
Irod =
1
m L2 ,
3
and the moment of inertia of the mass m with
respect to the pivot point is Imass = m L2 .
Thus the moment of inertia of the system is
I = Irod + Im
1
4
= m L2 + m L2 = m L2 .
3
3
010 (part 2 of 4) 10.0 points
The length C in the ﬁgure represents the location of the center-of-mass of the rod plus
mass system.
Determine the position of the center of mass
from the pivot point.
7
1. C = L
8
3
5. None of these
1
L
2
Explanation:
6. C =
1
The center of mass of the rod is L, so the
2
center of mass of the rod-plus-mass system is
1
L+L m
2
C=
m+m
1
1
3
= L+ L = L.
4
2
4
011 (part 3 of 4) 10.0 points
The unit is released from rest in the horizontal
position.
What is the kinetic energy of the unit when
the rod momentarily has a vertical orientation?
1
mgL
2
3
2. K = m g L correct
2
1. K =
3. K = 2 m g L
4. None of these
5. K =
5
mgL
2
6. K = m g L
Explanation:
The potential energy released can be computed in two ways:
∆U = ∆U
CM of rod
+ ∆U
mass
L
3
= mg + mgL = mgL,
2
2
or
2. C = L
3
L correct
4
5
4. C = L
8
3. C =
∆U |rod+m = ∆U |CM of rod+m
3
3
L = mgL.
= (2 m) g
4
2
xu (rx575) – HW08 – macdonald – (55475)
012 (part 4 of 4) 10.0 points
Let the kinetic energy be KV at the vertical
position and the moment of inertia of the
system be I.
Find the angular velocity of the system in
terms of KV and I.
2 KV
I
r
2 KV
2. ω =
correct
I
r
KV
3. ω =
I
1. ω =
4. None of these
4
ω2 r = α r
√
ω= α
and the angular velocity is
ω = ω0 + α ∆t
ω
ω − ω0
=
α=
∆t
∆t
since ω0 = 0, so
√
ω
α
1
t= =
=√
α
α
α
1
=p
= 0.451754 s .
KV
I
r
KV
6. ω =
2I
Explanation:
The rotational kinetic energy of the system
is given by
1
KV = I ω 2
2
r
2 KV
.
ω=
I
5. ω =
013 10.0 points
A horizontal disk with a radius of 20 m rotates about a vertical axis through its center.
The disk starts from rest and has a constant
angular acceleration of 4.9 rad/s2 .
At what time will the radial and tangential components of the linear acceleration of
a point on the rim of the disk be equal in
magnitude?
Explanation:
Let :
r = 20 m and
The tangential and radial accelerations are
equal:
ar = at
014 (part 1 of 2) 10.0 points
Consider a 5 kg square which has its mass
concentrated along its perimeter, with each
side of length 6 m.
d
d
What is the moment of inertia of the square
about an axis perpendicular to the plane of
the square at its center of mass? Use the
parallel axis theorem and divide the square
into parts. The moment of inertia of a rod
1
CM
m d2 .
rotated about its CM is Irod
=
12
Correct answer: 60 kg &middot; m2 .
Explanation:
Let :
5 kg
and
4
ℓ = 6 m.
mrod =
By the parallel axis theorem, the moment
of inertia of each rod about the center of the
square is
2
d
1
1
rod
2
= mrod d2 .
Isquare =
mrod d +mrod
12
2
3
xu (rx575) – HW08 – macdonald – (55475)
Since there are four sides to the square,
the moment of inertia of the square about its
center of mass is
1
CM
rod
2
Isquare = 4 Isquare = 4
mrod d
3
4 5 kg
4
) (6 m)2
= mrod ℓ2 = (
3
3
4
= 60 kg &middot; m2 .
015 (part 2 of 2) 10.0 points
d
d
What is the moment of inertia of the square
about an axis perpendicular to the plane of
the square at one of its corners?
Correct answer: 150 kg &middot; m2 .
Explanation:
The distance from the center of the square
d
to one of its corners is √ . Using the parallel
2
axis theorem, the moment of inertia of the
square about an axis at one of its corners is
corner
Isquare
2
1
d
2
=
msquare d + msquare √
3
2
1 1
=
msquare d2
+
3 2
5
= msquare d2
6
5
= (5 kg) (6 m)2
6
= 150 kg &middot; m2 .
016 10.0 points
Determine the moment of inertia of a cylinder
of radius 0.71 m, height 2 m and density
(1.18 − 0.59 r + 0.339 r 2 ) kg/m3 about the
center.
Correct answer: 0.765446 kg &middot; m2 .
5
Explanation:
Let :
I=
Z
Z
r = 0.71 m
h = 2 m.
and
ρ r 2 dV
R
(1.18 − 0.59 r + 0.339 r 2 ) r 2 r dr
0
Z 2π Z h
&times;
dθ
dz
0
0
R5
R6
R4
2πh
− 0.59
+ 0.339
= 1.18
4
5
6
(0.71 m)4
(0.71 m)5
= 1.18
− 0.59
4
5
(0.71 m)6
2 π (2 m)
+ 0.339
6
=
= 0.765446 kg &middot; m2 .
017 10.0 points
The tub of a washer goes into its spin-dry
cycle, starting from rest and reaching an angular speed of 5.5 rev/s in 7.7 s . At this point
the person doing the laundry opens the lid,
and a safety switch turns oﬀ the washer. The
tub slows to rest in 13.7 s .
During the whole process of starting up
and stopping, through how many revolutions
does the tub turn? Assume constant angular
acceleration while it is starting and stopping.
Explanation:
Let’s split the problem into two pieces, the
ﬁrst phase as the washer is starting to spin,
and the second phase while it is slowing down.
In both cases, we will use our two equations
of angular motion:
ωf = ω0 + α t
1
θ f = θ 0 + ω0 t + α t2 .
2
Now, during the ﬁrst phase of the motion
(when the washer is starting up), its initial angular velocity is zero, ω0 = 0, its ﬁnal angular
xu (rx575) – HW08 – macdonald – (55475)
velocity is 5.5 rev/s, and the time it takes to
reach this angular velocity is 7.7 s. We can
solve for the angular acceleration using the
ﬁrst equation above:
ωf = ω0 + α t
5.5 rev/s = α(7.7 s)
5.5 rev/s
⇒α=
= 0.714286 rev/s2 .
7.7 s
Now we can use the angular position equation to determine the number of revolutions
during the ﬁrst phase of the motion, letting
θ0 = 0:
1
θ f = θ 0 + ω0 t + α t2
2
1
θf = (0.714286 rev/s2 )(7.7 s)2
2
⇒ θf = 21.175 rev .
Now we repeat the process for the second
phase of the motion, where now the initial
angular velocity is ω0 = 5.5 rev/s and the
ﬁnal angular velocity is ωf = 0. The time it
takes to decelerate is t = 13.7 s. We again
solve for the angular deceleration using the
angular velocity equation:
ωf = ω0 + α t
0 = 5.5 rev/s + α(13.7 s)
5.5 rev/s
⇒α=−
= −0.40146 rev/s2 .
13.7 s
Now we again use the angular position
equation to determine the number of revolutions during the second phase of the motion,
letting θ0 = 21.175 rev:
1
θ f = θ 0 + ω0 t + α t2
2
θf = 21.175 rev + 5.5 rev/s(13.7 s)
1
− (0.40146 rev/s2 )(13.7 s)2
2
⇒ θf = 58.85 rev .
Since we included the number of revolutions made during the ﬁrst phase as θ0 in our
calculation for the second phase, this is the
total number of revolutions.
6
018 (part 1 of 3) 10.0 points
What is the tangential acceleration of a bug
on the rim of a 78 rpm record of diameter
14.42 in. if the record moves from rest to its
ﬁnal angular speed in 4 s? The conversion
between inches and meters is 0.0254 m/in.
Explanation:
Let :
ω = 78 rpm ,
r = (7.21 in) (0.0254 m/in) = 0.183134 m
t = 4 s.
ω = ω0 + α t = α t
∆ω
α=
, so
∆t
r ∆ω
∆t
(0.183134 m) (78 rpm) 2 π 1 min
=
4s
1 rev 60 s
at = r α =
= 0.373966 m/s2 .
019 (part 2 of 3) 10.0 points
When the record is at its ﬁnal speed, what is
the tangential velocity of the bug?
Explanation:
vt = r ω
= (0.183134 m)(78 rpm)
2 π 1 min
1 rev 60 s
= 1.49586 m/s .
020 (part 3 of 3) 10.0 points
What is its radial acceleration 2.6 s after starting from rest? Assume that the record has
constant angular acceleration.
xu (rx575) – HW08 – macdonald – (55475)
Each spoke contributes to the total moment
of inertia as a thin rod pivoted at one end, so
2
Explanation:
Since the record has constant angular acceleration, and starts from rest we know that
ωf
ω = αt =
t.
tf
Knowing this, we know that the relationship
between radial acceleration and the velocity
of rotation is given by:
2
ωf
v2
2
ar = = ω r = t
r
r
tf
2
78 rpm 2 π
0.183134 m
= (2.6 s)
s
4 s 60 min
2
= 5.16229 m/s .
m R2
I = Ihoop + n Irod = M R2 + n
3
m 2
R
= M +n
3
0.087 kg
= 1 kg + 7
(0.62 m)2
3
= 0.462433 kg &middot; m2 .
022 (part 2 of 2) 10.0 points
Determine the moment of inertia of the wheel
about an axis through its rim and perpendicular to the plane of the wheel.
Correct answer: 1.08093 kg &middot; m2 .
Explanation:
From the parallel axis theorem
021 (part 1 of 2) 10.0 points
A wheel is formed from a hoop of mass 1 kg
and seven equally spaced spokes, each of mass
0.087 kg. The hoop’s radius is the length
0.62 m of each spoke.
0.62 m
0.087 kg
Irim = I + (M + n m) R2
= 0.462433 kg &middot; m2
+ [1 kg + 7 (0.087 kg)] (0.62 m)2
= 1.08093 kg &middot; m2 .
023 10.0 points
Find the moment of inertia of a solid sphere
tangent to the sphere.
m
1 kg
Find the moment of inertia of the wheel
about an axis through its center and perpendicular to the plane of the wheel.
Correct answer: 0.462433 kg &middot; m2 .
Explanation:
Let :
7
n = 7,
m = 0.087 kg ,
M = 1 kg , and
R = 0.62 m .
ω
8
M R2
5
4
2. I = M R2
5
3
3. I = M R2
5
9
4. I = M R2
5
1. I =
r
xu (rx575) – HW08 – macdonald – (55475)
7
M R2 correct
5
6
6. I = M R2
5
Explanation:
The moment of inertia of a solid about a
diameter is
8
N
5. I =
Icm =
θ
sin
g
m
P
θ
mg
2
M R2 .
5
Using the parallel-axis theorem, the moment
of inertia about an axis that is tangent to the
sphere is
The moment of inertia of the ball about
the point of contact between the ball and the
inclined plane is
I = Icm + M R2
2
= M R2 + M R2
5
7
= M R2 .
5
IP = Icm + m d2
2
= m R2 + m R2
3
5
= m R2 .
3
024 10.0 points
A 140 g basketball has a 37.7 cm diameter
and may be approximated as a thin spherical
shell.
The net torque about the point of contact
between the ball and the inclined plane is
5. 3
9m
5
m R2 α
3
3
m g R2 sin θ
= g sin θ .
a=
5
5
m R2
3
m g R sin θ = IP α =
140 g
0
&micro;=
. 37
25◦
Starting from rest, how long will it take
a basketball to roll without slipping 5.39 m
down an incline that makes an angle of 25 ◦
with the horizontal? The moment of inertia of
a thin spherical shell of radius R and mass m
2
is I = m R2 , the acceleration due to gravity
3
is 9.8 m/s2 , and the coeﬃcient of friction is
0.37 .
Explanation:
Let : ℓ = 5.39 m ,
θ = 25◦ , and
R = 0.1885 m .
Because the sphere starts from rest, its center of mass moves a distance
1 2
at
2
s
r
2ℓ
2 (5.39 m)
=
t=
a
2.485 m/s2
ℓ=
= 2.08279 s .
025 10.0 points
Consider a wheel (solid disk) of radius 1.41 m,
1
mass 15 kg and moment of inertia M R2 .
2
The wheel rolls without slipping in a straight
line in an uphill direction 16◦ above the horizontal. The wheel starts at angular speed
14.7518 rad/s but the rotation slows down
xu (rx575) – HW08 – macdonald – (55475)
as the wheel rolls uphill, and eventually the
wheel comes to a stop and rolls back downhill.
How far does the wheel roll in the uphill
direction before it stops? The acceleration of
gravity is 9.8 m/s2 .
9
3 R2 ω02
L=
4 g sin θ
=
4 (9.8 m/s2 ) sin 16◦
= 120.122 m .
Explanation:
Let : R = 1.41 m ,
M = 15 kg ,
θ = 16◦ ,
g = 9.8 m/s2 .
The actual value of the wheel’s mass is extraneous to this situation.
and
Since the wheel rolls without slipping, v =
R ω the friction force does no work so the net
mechanical energy of the wheel is conserved:
potential
kinetic
kinetic
Elinear
+ Erotation
+ Egravity
= const
1
1
M v 2 + I ω 2 + M g y = const
2
2
1
1 1
2
2
M (R ω) +
M R ω2 + M g y
2
2 2
= const
3
M R2 ω 2 + M g y = const .
4
The wheel starts at initial angular speed
ω = ω0 , but as it rolls uphill, its potential
energy increases while the kinetic energy decreases, and eventually the kinetic energy is
completely exhausted and the wheel stops.
By energy conservation, this happens when
1 2
ω + M g y0 = 0 + M g ystop .
4 0
During all this time, the wheel’s axis moves
some distance L in a straight line in the direction θ = 16 ◦ above the horizontal, so the
wheel’s elevation y increases by
∆y ≡ ystop − y0 = L sin θ .
Therefore,
3
M R2 ω02 = M g ∆y = M g L sin θ .
4
026 10.0 points
A uniform solid sphere with moment of inertia
2
m r 2 . mass 1.3 kg and radius 0.378 m is
5
placed on the inside surface of a hemispherical
bowl of radius 1.7 m. The sphere is released
from rest at an angle 69.2◦ from the vertical
and rolls without slipping.
R
θ
r
What is the angular velocity of the sphere
when it reaches the bottom of the bowl? The
acceleration of gravity is 9.8 m/s2 .
Explanation:
The distance between the center of mass of
the sphere at its highest (rest) and its lowest
points is
h = (R − r) − (R − r) cos θ
= (R − r)(1 − cos θ)
= (1.7 m − 0.378 m)(1 − cos 69.2◦ )
= 0.852549 m .
v = r ω , so from conservation of energy.
Ui = K f
1
m g h = m vf2 +
2
1
= m vf2 +
2
1
I ω2
2 f
vf 2
1 2
2
mr
2 5
r
xu (rx575) – HW08 – macdonald – (55475)
7
m vf2
10
r
10
vf =
gh
7
=
Thus,
vf
1
=
ωf =
r
r
r
10
gh
7
keywords:
10
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