madrid (tmm2353) – HW 113 – Antoniewicz – (56445)
1
This
printout
should
have
29
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001(part1of7)3.0points
A disk of radius 7
.
6 cm and mass 0
.
43 kg is
pulled along a frictionless surface with a force
of 14
.
1 N by a string wrapped around the edge
(Fig. 11.43, displayed below). 24 cm of string
has unwound off the disk.
At the instant
shown the angular velocity is 19 rad
/
s. What
is the magnitude of the torque exerted about
the center of the disc at this instant?
7
.
6 cm
14
.
1 N
24 cm
Correct answer: 1
.
0716 N
·
m.
Explanation:
Use
vector
τ
=
vectorr
×
vector
F
. Therefore,

vector
τ

=

vectorr
⊥

vector
F

=
(0
.
076 m) (14
.
1 N). Thus,

vector
τ

= 1
.
0716 N
·
m.
002(part2of7)2.0points
What is the direction of the torque exerted
about the center of the disc at this instant?
(choice of axes:
x
to the right,
y
up,
z
out of
the page)
1.
y
2.

y
3.

x
4.

z
correct
5.
z
6.
x
Explanation:
The direction of the torque, given by the
righthand rule, is in the negative
z
direction.
So,
vector
τ
=
<
0
,
0
,
−
1
.
0716 N
·
m
>
N
·
m.
003(part3of7)2.0points
At the same instant, what is the magnitude
of the angular momentum about the center of
the disc?
Correct answer: 0
.
023595 kg
·
m
2
/
s.
Explanation:
vector
L
=
Ivectorω
=
<
0
,
0
,
−
Iω
>
,
where
I
=
(1
/
2)
MR
2
for the solid disc and
vectorω
is directed
along the negative
z
axis.
Using the values
given from the beginning of the problem, the
equation becomes
vector
L
=
<
0
,
0
,
−
(1
/
2)
MR
2
ω >
=
<
0
,
0
,
−
0
.
023595 kg
·
m
2
/
s
>

vector
L

= 0
.
023595 kg
·
m
2
/
s
004(part4of7)2.0points
At the same instant, what is the direction
angular momentum about the center of the
disc?
1.
z
2.

y
3.
y
4.

z
correct
5.
x
6.

x
Explanation:
The direction of the angular momentum is
the same as that of the angular velocity,
vectorω
.
005(part5of7)2.0points
At time Δ
t
= 0
.
2 s later, what is the direction
of the change of the angular momentum about
the center of the disk?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
madrid (tmm2353) – HW 113 – Antoniewicz – (56445)
2
1.
x
2.

x
3.

z
correct
4.
z
5.

y
6.
y
Explanation:
The angular momentum increases in the
direction of the angular velocity, so the the
change in angular momentum is in the nega
tive
z
direction.
006(part6of7)2.0points
At time Δ
t
= 0
.
2 s later, what is the mag
nitude of the angular momentum about the
center of the disk?
Correct answer: 0
.
237915 kg
·
m
2
/
s.
Explanation:
Since the torque acting on the disc is con
stant over time, we conclude the change in
angular momentum is given by
Δ
vector
L
=
vector
τ
Δ
t
Δ
vector
L
=
<
0
,
0
,
−
RF
Δ
t >
The final angular momentum is then given by
vector
L
f
=
vector
L
+ Δ
vector
L
=
vector
L
+
<
0
,
0
,
−
RF
Δ
t >
=
<
0
,
0
,
−
(1
/
2)
MR
2
ω
−
RF
Δ
t >
=
<
0
,
0
,
−
0
.
023595
−
0
.
21432
> kg
·
m
2
/s
=
<
0
,
0
,
−
0
.
237915 kg
·
m
2
/
s
>

vector
L
f

= 0
.
237915 kg
·
m
2
/
s
The change in angular momentum is in the
same direction as the initial angular momen
tum, so the magnitude of
vector
L
increases in the
negative
z
direction.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Turner
 Angular Momentum, Force, Correct Answer

Click to edit the document details