salazar (ggs332) – HW09 – TSOI – (58160)
1
This printout should have 27 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A circularshaped object oF mass 11 kg has
an inner radius oF 15 cm and an outer radius
oF 21 cm. Three Forces (acting perpendicular
to the axis oF rotation) oF magnitudes 12 N,
27 N, and 14 N act on the object, as shown.
The Force oF magnitude 27 N acts 21
◦
below
the horizontal.
12 N
14 N
27 N
21
◦
ω
±ind the magnitude oF the net torque on
the wheel about the axle through the center
oF the object.
Correct answer: 1
.
41 N
·
m.
Explanation:
Let :
a
= 15 cm = 0
.
15 m
,
b
= 21 cm = 0
.
21 m
,
F
1
= 12 N
,
F
2
= 27 N
,
F
3
= 14 N
,
and
θ
= 21
◦
.
F
1
F
3
F
2
θ
ω
The total torque is
τ
=
a F
2

b F
1

b F
3
= (0
.
15 m) (27 N)

(0
.
21 m) (12 N + 14 N)
=

1
.
41 N
·
m
,
with a magnitude oF
1
.
41 N
·
m
.
002
(part 1 oF 2) 10.0 points
A particle is located at the vector position
vr
= (1
.
7 m)ˆ
ı
+ (3
.
6 m)ˆ
and the Force acting on it is
v
F
= (2
.
9 N)ˆ
ı
+ (2
.
3 N)ˆ
.
What is the magnitude oF the torque about
the origin?
Correct answer: 6
.
53 N m.
Explanation:
Basic Concept:
v
τ
=
×
v
F
Solution:
Since neither position oF the par
ticle, nor the Force acting on the particle have
the
z
components, the torque acting on the
particle has only
z
component:
v
τ
= [
x F
y

y F
x
]
ˆ
k
= [(1
.
7 m) (2
.
3 N)

(3
.
6 m) (2
.
9 N)]
ˆ
k
= [

6
.
53 N m]
ˆ
k .
003
(part 2 oF 2) 10.0 points
What is the magnitude oF the torque
about the point having coordinates [
a, b
] =
[(0
.
8 m)
,
(7 m)]?
Correct answer: 11
.
93 N m.
Explanation:
Reasoning similarly as we did in the pre
vious section, but with the di²erence that
relative to the point [(0
.
8 m)
,
(7 m)] the
y

component oF the particle is now [
y

(7 m)],
we have
v
τ
=
{
[
x

a
]
F
y

[
y

b
]
F
x
}
ˆ
k
=
{
[(1
.
7 m)

(0
.
8 m)] [2
.
3 N]

[(3
.
6 m)

(7 m)] [2
.
9 N]
}
ˆ
k
=
{
11
.
93 N m
}
ˆ
k .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentsalazar (ggs332) – HW09 – TSOI – (58160)
2
004
10.0 points
A uniform 1 kg rod with length 4 m has
a frictionless pivot at one end. The rod is
released from rest at an angle of 21
◦
beneath
the horizontal.
2 m
4 m
1 kg
21
◦
What is the angular acceleration of the rod
immediately after it is released? The moment
of inertia of a rod about the center of mass
is
1
12
mL
2
,
where
m
is the mass of the rod
and
L
is the length of the rod. The moment
of inertia of a rod about either end is
1
3
2
,
and the acceleration of gravity is 9
.
8 m
/
s
2
.
Correct answer: 3
.
43091 rad
/
s
2
.
Explanation:
Let :
m
= 1 kg
,
L
= 4 m
,
and
θ
= 21
◦
.
The rod’s moment of inertia about its end
point is
I
=
1
3
2
,
so the angular accelera
tion of the rod is
α
=
τ
I
=
1
2
mg L
cos
θ
1
3
2
=
3
2
g
L
cos
θ
=
3
2
9
.
8 m
/
s
2
4 m
cos 21
◦
=
3
.
43091 rad
/
s
2
.
005
10.0 points
Consider the setup shown, where the inclined
plane has a frictionless surface. The blocks
have masses
m
2
and
m
1
.
The pulley has
mass
m
3
, and is a uniform disc with radius
R
.
Assume the pulley to be frictionless.
m
1
T
1
R
m
3
m
2
μ
= 0
T
a
θ
What is the acceleration of the blocks? As
sume the mass
m
1
is more massive and is
descending with acceleration
a
. The moment
of inertia of a disk is
1
2
M R
2
. The accelera
tion of gravity is 9
.
8 m
/
s
2
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 TSOI
 Physics, Angular Momentum, Kinetic Energy, Mass, Moment Of Inertia, Rotation

Click to edit the document details