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Chapter 10
313
P10.74
Consider the total weight of each hand to act at the center of gravity (midpoint) of that hand. Then
the total torque (taking CCW as positive) of these hands about the center of the clock is given by
τθ
θ
=−
F
H
G
I
K
J
−
F
H
G
I
K
J
+
mg
L
L
g
mL
m L
h
h
hm
m
mh
h
h
m
m
m
22
2
sin
sin
sin
sin
bg
If we take
t
= 0 at 12 o’clock, then the angular positions of the hands at time
t
are
θω
hh
t
=
,
where
ω
π
h
=
6
rad h
and
mm
t
=
,
where
ωπ
m
=
2 r
adh
Therefore,
τ
F
H
G
I
K
J
+
L
N
M
O
Q
P
490
600
270
6
100
2
..
.
s
i
n
s
i
n
m s
kg
m
kg 4.50 m
2
af
t
t
or
⋅
F
H
G
I
K
J
+
L
N
M
O
Q
P
794
6
278
2
N m sin
.
sin
t
t
, where
t
is in hours.
(a)
(i)
At 3:00,
t
=
300
.
h
,
so
⋅
F
H
G
I
K
J
+
L
N
M
O
Q
P
⋅
794
2
6
794
Nm
sin
.
sin
(ii)
At 5:15,
t
=+
=
5
15
60
525
h
h
h
.
, and substitution gives:
⋅
2510 N m
(iii)
At 6:00,
=⋅
0 N m
(iv)
At 8:20,
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Gravity

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