In a simple pendulum the breaking strength of the string is double the weight of the bob the bob is released from the rest when the string as horizontal the string string breaks when it's make an angle theta with the vertical
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Explanation:
The correct option is C
θ
=
cos
−
1
(
2
3
)
Let the speed of bob at point B be
v
From the FBD:
T
−
m
g
cos
θ
=
m
v
2
l
…
(
i
)
(where
l
is the length of the string)
From conservation of energy,
m
g
l
=
m
g
l
(
1
−
cos
θ
)
+
1
2
m
v
2
{Bob is released from horizontal }
⇒
v
=
√
2
g
l
cos
θ
…
(
i
i
)
T
=
2
m
g
(Given)
⇒
2
m
g
=
m
g
cos
θ
+
m
v
2
l
(from (i))
⇒
m
v
2
=
2
m
g
l
−
m
g
l
cos
θ
⇒
m
(
2
g
l
cos
θ
)
=
2
m
g
l
−
m
g
l
cos
θ
(substituting from (ii))
⇒
3
m
g
l
cos
θ
=
2
m
g
l
⇒
cos
θ
=
2
3
∴
θ
=
cos
−
1
(
2
3
)
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