Question

Q.A pendulum consisting of a massless string of length 20 cm and a tiny bob
of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate kinetic energy and increase in
the gravitational potential energy of the bob. (Use π² = 10)


#QuaLiTy AnSWer neeDed
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Answer 1

Answer:

17. A pendulum consisting of a massless

string of length 20 cm and a tiny bob

of mass 100 g is set up as a conical

...

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Answer 2

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A pendulum consisting of a massless string of length l 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate kinetic energy and increase in the gravitational potential energy of the bob. (Use π² = 10).

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: firstly \: let's \: understand \: the \:question}}}$

In the question there is given that there is a massless string of 20 length and it contain a tiny Bob of mass 100 gram and it is set up as a conical pendulum. Its bob now performs 75 rpm (in min )

And we need to Calculate kinetic energy and increase in the gravitational potential energy of the bob.

★ (Use π² = 10)

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$\begin{gathered}\\\;\large{\boxed{\sf{ {T}^{} \;=\;\bf{2\pi \sqrt{ \dfrac{L \cos \theta}{g} } }}}}\end{gathered}$

$\begin{gathered}\\\;\large{\boxed{\sf{ {V}^{2} \;=\;\bf{rg \tan \theta } }}}\end{gathered}$

$\begin{gathered}\\\;\large{\boxed{\sf{ΔPE \;=\;\bf{mg(L - h)} }}}\end{gathered}$

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~ Let's write what is given to us and what we need to solve

⌬ Length of string (L) = 20cm = 0.2m

⌬ Mass of Bob (m) = 100g = 0.1 kg

⌬ Revolution per sec (n) = 75rpm = 5/4rms

⌬ Gravity (g) = 10 m/s² [ π² = 10 ]

⌬ Time period (T) = ¹/n = ⅘ = 0.8 s

$\begin{gathered}\\\;\large{\boxed{\sf{ {T}^{} \;=\;\bf{2\pi \sqrt{ \dfrac{L \cos \theta}{g} } }}}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ {T}^{2} \;=\;\bf{ {4}^{2} \pi { \dfrac{L \cos \theta}{g} } }}}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ height\;=\;\bf{{{L \cos \theta = \frac{9T^{2} }{ {4\pi}^{2} } }} }}}}\end{gathered}$

~ Put all the values ....!

• T = 0.8

• π = 10

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ height\;=\;\bf{{{ \frac{(10) \times (0.8) }{ {4 \times (10)} } }} }}}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ height\;=\;\bf{{{ {0.16m} }} }}}}\end{gathered}$

So , value of cos theta

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ \cos\theta\;=\;\bf{{{ { \frac{0.16}{0.2 } = 0.8 } }} }}}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf \therefore{ \theta\;= \cos ^{ - 1} 0.8 = \;\bf{{{ { 36 .87 \degree = 36 \degree5' } }} }}}}\end{gathered}$

$\begin{gathered}\\\;\large{\boxed{\sf{ {V}^{2} \;=\;\bf{rg \tan \theta } }}}\end{gathered}$

~ Put all the values ....!

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ {V}^{2} \;=\;\bf{(L \sin \theta)(g)( \tan36 \degree5')} }}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ \;\bf{0.9} }}}\end{gathered}$

The kinetic energy of Bob = ½ mv²

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ \;\bf{ \frac{1}{2} } \: ( {0.1})(0.9) }}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ \;\bf{ 0.045 \: J} }}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\\;\underline{\boxed{\tt{\odot\;\;Hence,\;\; kinetic\;energy=\;\bf{\blue{0.045 \: J}}}}}\end{gathered} \end{gathered}\end{gathered}$

the increase in gravitational potential energy of the bob

$\begin{gathered}\\\;\large{\boxed{\sf{ΔPE \;=\;\bf{mg(L - h)} }}}\end{gathered}$

$\begin{gathered}\\\;\large{\longrightarrow{\sf{ΔPE \;=\;\bf{(0.1)(10)(0.2- 0.16)} }}}\end{gathered}$

$\begin{gathered}\\\;\large{\Longrightarrow{\sf{ΔPE \;=\;\bf{0.04 \: J} }}}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\\;\underline{\boxed{\tt{\odot\;\;Hence,\;\;gravitational \;potential \;energy \;=\;\bf{\blue{0.04 \: J}}}}}\end{gathered} \end{gathered}\end{gathered}$

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: know \: more}}}$

What is conical pendulum ?

• It consists of a bob fixed on the end of a string suspended from a pivot.

• The bob of a conical pendulum moves at a constant speed in a circle with the string tracing out a cone.