Answer
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Hint:The relationship between the moment of inertia and angular velocity can help to solve this problem. Also the law of conservation of the momentum can be used to solve this problem as the problem involves rotations and it is given that the moment of inertia of the dancer changes as she folded her arms and the angular momentum involves the terms angular velocity and moment of inertia.
Formula used: The formula of the angular momentum is given by $L = I \cdot \omega $ where $L$ is the angular momentum, $I$ is the moment of inertia and $\omega $ is the angular velocity.
Complete step by step answer:
As it is given as the initial, the dancer has 90 rpm when arms stretched and as the arms are folded the moment of inertia changes to $75\% $.
The angular momentum when arms were stretched is${L_1} = {I_1} \cdot {\omega _1}$ and after the arms are folded then the angular momentum is${L_2} = {I_2} \cdot {\omega _2}$. It is given that the moment of inertia changes to $75\% $ of its initial value therefore,
${L_2} = \dfrac{{75}}{{100}} \cdot {I_1} \cdot {\omega _2}$
According to the law of conservation of angular momentum.
$
{L_1} = {L_2} \\
\Rightarrow {I_1} \cdot {\omega _1} = \dfrac{{75}}{{100}} \cdot {I_1} \cdot {\omega _2} \\
\Rightarrow {\omega _1} = \dfrac{3}{4}{\omega _2} \\
\Rightarrow {\omega _2} = \dfrac{4}{3}{\omega _1} \\
\Rightarrow {N_2} = \dfrac{4}{3}{N_1} \\
$
It is given that the value of${N_1} = 90$.
${N_2} = \dfrac{4}{3}{N_1}$
Replace the value of${N_1} = 90{\text{ rpm}}$.
$
{N_2} = \dfrac{4}{3}{N_1} \\
\Rightarrow {N_2} = \left( {\dfrac{4}{3}} \right) \cdot \left( {90} \right) \\
\Rightarrow {N_2} = 120{\text{ rpm}} \\
$
The correct answer for this problem is ${N_2} = 120{\text{ rpm}}$. The new speed for the dancer is 120 rpm.
Note: The angular momentum is rotational momentum like linear momentum. The total angular momentum of a closed system remains constant; this is the law of conservation of angular momentum. It is advisable for students to remember the formulas of angular momentum to solve these types of problems.
Formula used: The formula of the angular momentum is given by $L = I \cdot \omega $ where $L$ is the angular momentum, $I$ is the moment of inertia and $\omega $ is the angular velocity.
Complete step by step answer:
As it is given as the initial, the dancer has 90 rpm when arms stretched and as the arms are folded the moment of inertia changes to $75\% $.
The angular momentum when arms were stretched is${L_1} = {I_1} \cdot {\omega _1}$ and after the arms are folded then the angular momentum is${L_2} = {I_2} \cdot {\omega _2}$. It is given that the moment of inertia changes to $75\% $ of its initial value therefore,
${L_2} = \dfrac{{75}}{{100}} \cdot {I_1} \cdot {\omega _2}$
According to the law of conservation of angular momentum.
$
{L_1} = {L_2} \\
\Rightarrow {I_1} \cdot {\omega _1} = \dfrac{{75}}{{100}} \cdot {I_1} \cdot {\omega _2} \\
\Rightarrow {\omega _1} = \dfrac{3}{4}{\omega _2} \\
\Rightarrow {\omega _2} = \dfrac{4}{3}{\omega _1} \\
\Rightarrow {N_2} = \dfrac{4}{3}{N_1} \\
$
It is given that the value of${N_1} = 90$.
${N_2} = \dfrac{4}{3}{N_1}$
Replace the value of${N_1} = 90{\text{ rpm}}$.
$
{N_2} = \dfrac{4}{3}{N_1} \\
\Rightarrow {N_2} = \left( {\dfrac{4}{3}} \right) \cdot \left( {90} \right) \\
\Rightarrow {N_2} = 120{\text{ rpm}} \\
$
The correct answer for this problem is ${N_2} = 120{\text{ rpm}}$. The new speed for the dancer is 120 rpm.
Note: The angular momentum is rotational momentum like linear momentum. The total angular momentum of a closed system remains constant; this is the law of conservation of angular momentum. It is advisable for students to remember the formulas of angular momentum to solve these types of problems.
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