garcia (jg46766) – Problem Set 10 – lacosse – (S2107)
1
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home stretch and hang in there!!
001(part1of2)10.0points
You are playing with a YoYo which you can
describe as a uniform disk with mass
m
and
radius
r
. The string of the YoYo has a length
L
. The YoYo rolls down vertically.
h
r
m
ω
Determine the tension
T
of the string at
a height Δ
h
below the original position
h
as
indicated in the sketch.
1.
T
=
1
3
m g
correct
2.
T
=
r
L
m g
3.
T
=
5
2
m g
4.
T
=
1
4
m g
5.
T
=
2
5
m g
6.
T
=
1
2
m g
7.
T
= 3
m g
8.
T
=
L

h
r
m g
9.
T
=
m g
10.
T
=
7
4
m g
Explanation:
Apply “
F
=
m a
” to the center of mass of
the disk,
F
net
=
summationdisplay
F
:
m g

T
=
m a .
(1)
Apply “
τ
=
I α
” to the YoYo (where the
torque is calculated about the center),
T r
=
I α
=
1
2
m r
2
α
=
1
2
m r a ,
so
(2)
T
=
1
2
m a .
(3)
In the 3
rd
step we used
a
=
α r
. Substituting
Eq. 3 into 1 leads to,
m g

1
2
m a
=
m a ,
so
a
=
2
3
g ,
and
T
=
1
2
m a
=
1
3
m g .
002(part2of2)10.0points
After the YoYo has been elastically reflected
at length
L
, it moves upward.
Determine the speed of its center when it is
at a height Δ
h
below the original position
h
.
1.
v
=
radicalbigg
r
L
g
Δ
h
2.
v
=
radicalbig
2
g
Δ
h
3.
v
=
radicalbigg
1
8
g
Δ
h
4.
v
=
radicalbig
g
Δ
h

g L
5.
v
=
radicalbigg
4
3
g
Δ
h
correct
6.
v
=
radicalbigg
3
2
g
Δ
h
7.
v
=
radicalbig
g
Δ
h
8.
v
=
radicalbigg
1
2
g
Δ
h
9.
v
=
radicalbigg
5
3
g
Δ
h
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garcia (jg46766) – Problem Set 10 – lacosse – (S2107)
2
10.
v
=
radicalbig
3
g
Δ
h
Explanation:
The application of workenergy equation
gives
m g
Δ
h
=
1
2
m v
2
+
1
2
I ω
2
.
But
1
2
I ω
2
=
1
2
parenleftbigg
1
2
m r
2
parenrightbigg
ω
2
=
1
4
m v
2
.
So
m g
Δ
h
=
parenleftbigg
1
2
+
1
4
parenrightbigg
m v
2
=
⇒
v
=
radicalbigg
4
3
g
Δ
h .
003(part1of2)10.0points
A particle is located at the vector position
vectorr
= (1
.
6 m)ˆ
ı
+ (4
.
2 m)ˆ
and the force acting on it is
vector
F
= (2
.
5 N)ˆ
ı
+ (3
.
2 N)ˆ
.
What is the magnitude of the torque about
the origin?
Correct answer: 5
.
38 N m.
Explanation:
BasicConcept:
vector
τ
=
vectorr
×
vector
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:
vector
τ
= [
x F
y

y F
x
]
ˆ
k
= [(1
.
6 m) (3
.
2 N)

(4
.
2 m) (2
.
5 N)]
ˆ
k
= [

5
.
38 N m]
ˆ
k .
004(part2of2)10.0points
What
is
the
magnitude
of
the
torque
about the point having coordinates [
a, b
] =
[(3 m)
,
(8 m)]?
Correct answer: 5
.
02 N m.
Explanation:
Reasoning similarly as we did in the pre
vious section, but with the difference that
relative to the point [(3 m)
,
(8 m)] the
y

component of the particle is now [
y

(8 m)],
we have
vector
τ
=
{
[
x

a
]
F
y

[
y

b
]
F
x
}
ˆ
k
=
{
[(1
.
6 m)

(3 m)] [3
.
2 N]

[(4
.
2 m)

(8 m)] [2
.
5 N]
}
ˆ
k
=
{
5
.
02 N m
}
ˆ
k .
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 Spring '11
 carter
 Physics, Angular Momentum, Force, Moment Of Inertia, Correct Answer

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