Explanation:
Apply
vector
L
≡
vectorr
×
vectorp
=
mvectorr
×
vectorv
vector
L
=
bracketleftBig
(
y
−
y
0
)ˆ
+ (
x
−
x
0
)ˆ
ı
bracketrightBig
×
m v
ˆ
ı
=
−
m v
(
y
−
y
0
)
ˆ
k
Then the magnitude of the angular momen
tum is
bardbl
vector
L
bardbl
=

m v
(
y
−
y
0
)

=

(2 kg)(7 m
/
s)(3 m
−
13 m)

= 140 kg m
2
/
s
006
10.0 points
A thin uniform cylindrical turntable of ra
dius 2
.
9 m and mass 28 kg rotates in a hori
zontal plane with an initial angular speed of
7
.
6 rad
/
s.
The turntable bearing is friction
less. A clump of clay of mass 11 kg is dropped
onto the turntable and sticks at a point 1
.
7 m
from the point of rotation.
Find the angular speed of the clay and
turntable.
Correct answer: 5
.
98424 rad
/
s.
Explanation:
Let :
R
= 2
.
9 m
,
x
= 1
.
7 m
,
M
= 28 kg
,
ω
= 7
.
6 rad
/
s
,
and
m
= 11 kg
.
Basic Concepts:
summationdisplay
vector
L
=
dvector
τ
ext
dt
Δ
K
=
K
f
−
K
i
=
Q .
From conservation of the angular momentum
it follows that
L
i
=
L
f
or
I
i
ω
=
I
f
ω
f
,
where
I
i
=
1
2
M R
2
=
1
2
(28 kg) (2
.
9 m)
2
= 117
.
74 kg m
2
.
Similarly,
I
f
=
I
i
+
m x
2
= (117
.
74 kg m
2
) + (11 kg) (1
.
7 m)
2
= 149
.
53 kg m
2
.
Finally,
ω
f
=
I
i
I
f
ω
=
(117
.
74 kg m
2
)
(149
.
53 kg m
2
)
(7
.
6 rad
/
s)
=
5
.
98424 rad
/
s
.
007
10.0 points
A projectile of mass
m
= 1
.
27 kg moves to
the right with speed
v
0
= 12
.
5 m
/
s. The pro
jectile strikes and sticks to the end of a sta
tionary rod of mass
M
= 8
.
65 kg and length
d
= 1
.
9 m that is pivoted about a frictionless
axle through its center.
I
cm rod
=
1
12
M L
2
d
O
O
v
0
ϖ
m
M
Find the angular speed of the system right
after the collision.
Correct answer: 4
.
02 rad
/
s.
Explanation:
kuruvila (lk5992) – HW 11 – opyrchal – (11113)
4
The initial angular momentum of the pro
jectile about the pivot O is
L
i
=
d
2
m v
0
. With
the projectile stuck to the end of the rod, the
rotational inertia of the projectile and the rod
combined about O is,
I
f
=
m
parenleftbigg
d
2
parenrightbigg
2
+
1
12
M d
2
=
bracketleftbigg
m
4
+
M
12
bracketrightbigg
d
2
(1)
=
bracketleftbigg
(1
.
27 kg)
4
+
(8
.
65 kg)
12
bracketrightbigg
(1
.
9 m)
2
= 3
.
75 kg m
2
.
Using conservation of angular momentum, we
have
L
i
=
I
f
ω
(2)
=
r
⊥
p
=
d
2
m v
0
(3)
Therefore, combining Eqs. (1), (2), and (3),
we have
ω
=
d
2
m v
0
m d
2
4
+
M d
2
12
·
12
12
=
6
m v
0
(3
m
+
M
)
d
=
6 (1
.
27 kg) (12
.
5 m
/
s)
[3 (1
.
27 kg) + (8
.
65 kg)] (1
.
9 m)
= 4
.
02 rad
/
s
.
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 Spring '08
 moro
 Angular Momentum, Mass, Moment Of Inertia, Rotation