Question

A ballet dancer revoled at 24 RPM will her hands folded if she stretched her hand so that her MI increases by 20% then new frequency of rotation will be​

Answer 1

$\bigstar\huge\boxed{\mathtt{\pink{Solution:}}}$

As per the given data ,

• Initial angular frequency of the ballet dancer with her hands folded = 24 rpm
• Let the moment of inertia initially be I
• The moment of inertia increases by 20 % when she streched her hands

Initially the moment of inertia was I  

Now it increases by 20 % so ,

$\implies\mathtt{I_{2}=I+\dfrac{20 I}{100} }$

$\implies\mathtt{I_{2}=I+0.2I }$

$\implies\mathtt{I_{2}= 1.2I }$

• Final moment of inertia = 1.2 I

Now we can easily find the final angular frequency  by using law of conservation of angular momentum,

$\implies\mathtt{L_{initial}= L_{final}}$

$\implies\mathtt{I_1\times\omega_1=I_2\times\omega_2}$

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we know that ,

$\star\:\mathtt{\omega=2\pi n}$

Now as the frequency is given in revolution per minute(rpm) we first need to convert it into revolution per second(rps)

In order to do that simply divide by 60

$\star\:\mathtt{\omega=\dfrac{2\pi n}{60}}$

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Now substituting the value of omega in the above equation we get ,

$\implies\mathtt{I_1\times\dfrac{2\pi n_1}{60}=I_2\times\dfrac{2\pi n_2}{60}}$

$\implies\mathtt{I_1 \times n_1=I_2\times n_2}$

$\implies\mathtt{I \times 24=1.2I\times n_2}$

$\implies\mathtt{n_2=\dfrac{24}{1.2} }$

$\implies\mathtt{\red{n_2=20 rpm }}$

The new frequency of rotation will be 20 rpm