
A light weight boy holds two heavy dumbbells of equal masses with outstretched arms while standing on a turn-table which is rotating with an angular frequency ${\omega _1}$ when the dumbbells are at distance ${r_1}$ from the axis of rotation. The boy suddenly pulls the dumbbells towards his chest until they are at distance ${r_2}$ from the axis of rotation. The new angular frequency of rotation ${\omega _2}$ of the turn-table will be equal to
A. ${\omega _1}\dfrac{{{r_2}}}{{{r_1}}}$
B. ${\omega _1}\dfrac{{{r_1}^2}}{{{r_2}^2}}$
C. ${\omega _1}\dfrac{{{r_1}}}{{{r_2}}}$
D. ${\omega _1}\dfrac{{{r_2}^2}}{{{r_1}^2}}$
Answer
499.2k+ views
Hint:In order to solve this question we need to understand conservation of angular momentum and energy. Angular momentum is defined as momentum in rotation and it is the quantity of rotation of the body and it is mathematically defined as the product of moment of inertia and its angular velocity. Angular momentum is conserved and torque is defined as the space derivative of angular momentum, so if a torque during rotation is zero then the space derivative of angular momentum is zero and hence angular momentum is constant or conserved during rotation.
Complete step by step answer:
Mathematically angular momentum is defined as, $J = I\omega $.
Here, $I$ is the moment of inertia and it is defined as $I = m{r^2}$.
So when dumbbells are far away from his chest then angular momentum is,
${J_1} = m{r_1}^2{\omega _1}$
Here, ${r_1}$ is radius and ${\omega _1}$ is angular frequency.
Similarly angular momentum when dumbbells are close to his chest is given by,
${J_2} = m{r_2}^2{\omega _2}$
Here, ${r_2}$ is radius and ${\omega _2}$ is angular frequency
Since angular momentum is conserved so ${J_1} = {J_2}$.
Putting values we get,
$m{r_1}^2{\omega _1} = m{r_2}^2{\omega _2}$
$\Rightarrow { r_1}^2{\omega _1} = {r_2}^2{\omega _2}$
$\therefore {\omega _2} = {\omega _1}\dfrac{{{r_1}^2}}{{{r_2}^2}}$
So the correct option is B.
Note: It should be remembered that, Due to law of conservation of angular momentum when the distance from axis of rotation gets decreased it increases the angular velocity of the body that’s why when a dumbbells are far away from chest angular velocity is less as compared to when dumbbells are closed to chest.
Complete step by step answer:
Mathematically angular momentum is defined as, $J = I\omega $.
Here, $I$ is the moment of inertia and it is defined as $I = m{r^2}$.
So when dumbbells are far away from his chest then angular momentum is,
${J_1} = m{r_1}^2{\omega _1}$
Here, ${r_1}$ is radius and ${\omega _1}$ is angular frequency.
Similarly angular momentum when dumbbells are close to his chest is given by,
${J_2} = m{r_2}^2{\omega _2}$
Here, ${r_2}$ is radius and ${\omega _2}$ is angular frequency
Since angular momentum is conserved so ${J_1} = {J_2}$.
Putting values we get,
$m{r_1}^2{\omega _1} = m{r_2}^2{\omega _2}$
$\Rightarrow { r_1}^2{\omega _1} = {r_2}^2{\omega _2}$
$\therefore {\omega _2} = {\omega _1}\dfrac{{{r_1}^2}}{{{r_2}^2}}$
So the correct option is B.
Note: It should be remembered that, Due to law of conservation of angular momentum when the distance from axis of rotation gets decreased it increases the angular velocity of the body that’s why when a dumbbells are far away from chest angular velocity is less as compared to when dumbbells are closed to chest.
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