lim (kl9356) – Homework 15 – Weathers – (17104)
1
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001
(part 1 of 2) 10.0 points
The hour and minute hands of Big Ben in
London are 2
.
65 m and 4
.
47 m long and have
masses of 57
.
6 kg and 95 kg respectively.
Calculate the total angular momentum of
the minute
hand
about the
center
point.
Treat the hand as long, thin rod. Treat “into
the clock” as the positive direction.
Correct answer: 1
.
10432 kg
·
m
2
/
s.
Explanation:
Basic Concepts:
L
=
I ω
Moment
of inertia of a
thin
rod
pivoted
through one of its ends is
I
=
m l
2
3
, where
m
is its mass and
l
is its length.
Solution:
Applying this formula to the
minute hand we obtain its moment of inertia:
I
min
=
m
min
l
2
min
3
=
(95 kg)(4
.
47 m)
2
3
= 632
.
728 kg
·
m
2
The angular momentum is
L
min
=
I
min
ω
min
=
(
632
.
728 kg
·
m
2
)
(0
.
00174533 1
/
s)
= 1
.
10432 kg
·
m
2
/
s
002
(part 2 of 2) 10.0 points
Calculate the total angular momentum of the
hour hand about the center point. Treat the
hand as long, thin rod, and “into the clock”
as the positive direction.
Correct answer: 0
.
0196105 kg
·
m
2
/
s.
Explanation:
This part is to be solved similarly to the previ
ous part. We obtain for the moment of inertia
of the hour hand:
I
hr
=
m
hr
l
2
hr
3
=
(57
.
6 kg)(2
.
65 m)
2
3
= 134
.
832 kg
·
m
2
The angular momentum is
L
hr
=
I
hr
ω
hr
= (134
.
832 kg
·
m
2
)(0
.
000145444 1
/
s)
= 0
.
0196105 kg
·
m
2
/
s
003
10.0 points
A small metallic bob is suspended from the
ceiling by a thread of negligible mass.
The
ball is then set in motion in a horizontal circle
so that the thread describes a cone.
The acceleration of gravity is 9
.
8 m
/
s
2
.
v
9
.
8 m
/
s
2
1
.
5 m
9 kg
31
◦
Calculate the magnitude of the angular
momentum of the bob about the support
ing point.
Correct answer: 14
.
8299 kg m
2
/
s.
Explanation:
Basic Concepts:
Angular Momentum
L
=
m vectorr
×
vectorv .
Note:
In this problem
vectorr
and
vectorv
are perpendic
ular, where
r
=
ℓ
sin
θ .
Let :
ℓ
= 1
.
5 m
,
θ
= 31
◦
,
g
= 9
.
8 m
/
s
2
,
and
m
= 9 kg
.
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lim (kl9356) – Homework 15 – Weathers – (17104)
2
Use the free body diagram below.
T
m g
θ
Solution:
The second Newton’s law in the
vertical and horizontal projections, respec
tively, in our case reads
T
cos
θ

m g
= 0
T
sin
θ

m ω
2
ℓ
sin
θ
= 0
,
where
T
is the force with which the wire acts
on the bob and the radius of the orbit is
R
=
ℓ
sin
θ
. From this system of equations we
find
ω
=
radicalbigg
g
ℓ
cos
θ
=
radicalBigg
(9
.
8 m
/
s
2
)
(1
.
5 m) cos(31
◦
)
= 2
.
7608 rad
/
s
.
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 Fall '08
 Weathers
 Magnetism, Angular Momentum, Mass, Work, lim

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