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F = Ma
2013 F = Ma
20
20
Part of
2013 F = Ma
Problems
(1)
2013 F=ma Problem 20
Source:
3/4/2013
A simple pendulum experiment is constructed from a point mass
m
m
m
attached to a pivot by a massless rod of length
L
L
L
in a constant gravitational field. The rod is released from an angle
θ
0
<
π
2
\theta_0 < \frac{\pi}{2}
θ
0
<
2
π
at rest and the period of motion is found to be
T
0
T_0
T
0
. Ignore air resistance and friction.What is the maximum value of the tension in the rod?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
m
g
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
m
g
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
m
L
θ
0
/
T
0
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
m
g
sin
θ
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
m
g
(
3
−
2
cos
θ
0
)
<span class='latex-bold'>(A) </span> mg\\ <span class='latex-bold'>(B) </span> 2mg\\ <span class='latex-bold'>(C) </span> mL\theta_0/T_0^2\\ <span class='latex-bold'>(D) </span> mg \sin \theta_0\\ <span class='latex-bold'>(E) </span> mg(3 - 2 \cos \theta_0)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
m
g
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
m
g
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
m
L
θ
0
/
T
0
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
m
g
sin
θ
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
m
g
(
3
−
2
cos
θ
0
)
trigonometry