What is a conical pendulum? Show that its
I cose
time periodisgiven by 210 where!
g
is the length of the string, 0 is the angle
that the string makes with the vertical and
g is the acceleration due to gravity.
Answers
Answer:
Conical Pendulum:-
Conical pendulum is defined as ,
If a mass or bob is tied to a string, and is rotated about a fixed axis Horizontally, I.e. in Horizontal circle is called a conical pendulum.
\rule{300}{1.5}
\rule{300}{1.5}
From Figure,
We can Make out The Tension in the string is "F".
Now,
As we can make out,
The Tension's is resolved in to components,
It's one component is Fsinθ and Fcosθ.
The Fsinθ Component gives the Centripetal Force,
Therefore,
\large{\boxed{\tt F = \dfrac{mv^2}{r}}}
F=
r
mv
2
Substituting the values,
\large{\tt \leadsto Fsin \theta = \dfrac{mv^2}{r}}⇝Fsinθ=
r
mv
2
\large{\tt \leadsto Fsin \theta = \dfrac{m(r \omega)^2}{r} \: \: \: \: \rightarrow\: \: \: \: (v = r \omega)}⇝Fsinθ=
r
m(rω)
2
→(v=rω)
Now,
\large{\tt \leadsto Fsin\theta = \dfrac{mr^2 \omega^2}{r}}⇝Fsinθ=
r
mr
2
ω
2
\large{\tt \leadsto Fsin\theta = \dfrac{m\cancel{r^2} \omega^2}{\cancel{r}}}⇝Fsinθ=
r
m
r
2
ω
2
\large{\underline{\underline{\tt Fsin \theta = m\omega^2 r}}}
Fsinθ=mω
2
r
Now, From the Figure we can see,
Applying Trigonometric ratios,
\large{\boxed{\tt Sin \theta = \dfrac{Opposite \: side}{Hypotenuse}}}
Sinθ=
Hypotenuse
Oppositeside
Opposite side = r
Hypotenuse = L.
Substituting the values,
\large{\tt \leadsto sin \theta = \dfrac{r}{L}}⇝sinθ=
L
r
Substituting in the above equation, we get,
\large{\tt \leadsto F \times \dfrac{r}{L} = m\omega^2r}⇝F×
L
r
=mω
2
r
\large{\tt \leadsto F \times \dfrac{\cancel{r}}{L} = m\omega^2\cancel{r}}⇝F×
L
r
=mω
2
r
\large{\tt \leadsto F \times \dfrac{1}{L} = m\omega^2}⇝F×
L
1
=mω
2
\large{\tt \leadsto F = m \omega^2 L \: -----(1)}⇝F=mω
2
L−−−−−(1)
\large{\boxed{\tt F = m\omega^2 L}}
F=mω
2
L
Now, From F.B.D of the mass "m",
\large{\tt Fcos \theta = mg}Fcosθ=mg
As both are antagonistic to each other,
Now, Substituting the values from equation (1),
\large{\tt \leadsto m \omega^2 L \times cos \theta = mg}⇝mω
2
L×cosθ=mg
\large{\tt \leadsto \cancel{m} \omega^2 L \times cos \theta = \cancel{m}g}⇝
m
ω
2
L×cosθ=
m
g
\large{\tt \leadsto \omega^2 \times L cos \theta = g}⇝ω
2
×Lcosθ=g
\large{\tt \leadsto \omega^2 = \dfrac{g}{L cos \theta}}⇝ω
2
=
Lcosθ
g
\large{\tt \leadsto \omega = \sqrt{\dfrac{g}{L cos \theta}}}⇝ω=
Lcosθ
g
As we know,
∵ω = 2π/T
Substituting this, we get,
\large{\tt \leadsto \dfrac{2 \pi}{T} = \sqrt{\dfrac{g}{L cos \theta}}}⇝
T
2π
=
Lcosθ
g
Now,Reciprocating it.
\large{\tt \leadsto \dfrac{T}{2 \pi} = \sqrt{\dfrac{L cos \theta}{g}}}⇝
2π
T
=
g
Lcosθ
\huge{\boxed{\boxed{T = 2 \pi \sqrt{\dfrac{L cos \theta}{g}}}}}
T=2π
g
Lcosθ
Explanation:
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