patel (kp9583) – HW9 – Mackie – (20211)
1
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001
(part 1 of 4) 10.0 points
An amusement park builds a ride in which
the victim is made to spin about a pole with
a rocket strapped on his seat. The rider, box
and rocket have an initial total mass of 240 kg.
Neglect the mass of the rod of length 4 m
.
260 m
/
s
75 N
4 m
240 kg
What is the moment of inertia of rider, box
and rocket about the pole? The acceleration
of gravity is 9
.
8 m
/
s
2
.
Treat the rider, box
and rocket as a point mass.
Correct answer: 3840 kg
·
m
2
.
Explanation:
Let :
F
= 75 N
,
v
t
= 260 m
/
s
,
M
= 240 kg
,
and
ℓ
= 4 m
.
For a point mass, the inertia is
I
=
M ℓ
2
= (240 kg) (4 m)
2
=
3840 kg
·
m
2
.
002
(part 2 of 4) 10.0 points
The rocket develops a thrust of 75 N perpen
dicular to the path of the rider.
What is the initial angular acceleration of
the rider?
Correct answer: 0
.
078125 rad
/
s
2
.
Explanation:
The torque of this force is
τ
=
ℓ F
=
Iα
α
=
ℓ F
I
=
(4 m) (75 N)
3840 kg
·
m
2
=
0
.
078125 rad
/
s
2
.
003
(part 3 of 4) 10.0 points
After what time
t
is the rider’s velocity equal
to 7 m
/
s? Neglect the change in mass of the
rocket.
Correct answer: 22
.
4 s.
Explanation:
Let :
v
= 7 m
/
s
.
ω
=
α t
for constant angular acceleration,
so
v
=
ℓ ω
=
ℓ α t
t
=
v
ℓ α
=
7 m
/
s
(4 m) (0
.
078125 rad
/
s
2
)
=
22
.
4 s
.
004
(part 4 of 4) 10.0 points
Gas exits the rocket at
v
t
= 260 m
/
s.
What mass per second must leave to de
velop the thrust
F
given above?
Correct answer: 0
.
288462 kg
/
s.
Explanation:
A flow of mass
d m
dt
with a velocity
v
gener
ates a thrust of
F
=
d m
dt
v
, so
d m
dt
=
F
v
t
=
75 N
260 m
/
s
=
0
.
288462 kg
/
s
.
005
10.0 points
An Atwood machine is constructed using a
hoop with spokes of negligible mass. The 2 kg
mass of the pulley is concentrated on its rim,
which is a distance 24
.
9 cm from the axle.
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patel (kp9583) – HW9 – Mackie – (20211)
2
The mass on the right is 1
.
03 kg and on the
left is 1
.
51 kg.
3
.
8 m
2 kg
24
.
9 cm
ω
1
.
51 kg
1
.
03 kg
What is the magnitude of the linear acceler
ation of the hanging masses? The acceleration
of gravity is 9
.
8 m
/
s
2
.
Correct answer: 1
.
03612 m
/
s
2
.
Explanation:
Let :
M
= 2 kg
,
R
= 24
.
9 cm
,
m
1
= 1
.
03 kg
,
m
2
= 1
.
51 kg
,
h
= 3
.
8 m
,
and
v
=
ω R .
Consider the free body diagrams
1
.
51 kg
1
.
03 kg
T
2
T
1
m
2
g
m
1
g
a
a
The net acceleration
a
=
r α
is in the direc
tion of the heavier mass
m
2
.
For the mass
m
1
,
F
net
=
m
1
a
=
m
1
g

T
1
T
1
=
m
1
g

m
1
a
and for the mass
m
2
,
F
net
=
m
2
a
=
T
2

m
2
g
T
2
=
m
2
a
+
m
2
g .
The pulley’s mass is concentrated on the
rim, so
I
=
M r
2
,
and
τ
net
=
summationdisplay
τ
ccw

summationdisplay
τ
cw
=
I α
T
1
r

T
2
r
= (
m r
2
)
parenleftBig
a
r
parenrightBig
=
m r a
m a
=
T
1

T
2
m a
= (
m
1

m
2
)
g

(
m
1
+
m
2
)
a
m a
+ (
m
1
+
m
2
)
a
= (
m
1

m
2
)
g
a
=
(
m
1

m
2
)
g
m
+
m
1
+
m
2
=
(1
.
03 kg

1
.
51 kg) (9
.
8 m
/
s
2
)
2 kg + 1
.
03 kg + 1
.
51 kg
=
1
.
03612 m
/
s
2
.
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 Fall '09
 MIHALISIN
 Physics, Angular Momentum, Mass, Rotation, Correct Answer, patel

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