
The principle involved in the performance of a spinning chair circus acrobat is
A. Conservation of angular momentum
B. Conservation of linear momentum
C. Conservation of energy
D. Principle of moment
E. Work energy principle
Answer
577.5k+ views
Hint
Firstly, we will write down the formula of torque in terms of moment of inertia and angular acceleration. We know that the rate of change of angular momentum equals the torque. Here the external torque is zero. So, we know that the angular momentum remains conserved. So, we will calculate that when an acrobat curls his body and brings his hands and legs as close to his body then his moment of inertia decreases and consequently his angular velocity increases. So, we can say easily that the principle involved is conservation of angular momentum.
Complete step by step answer
If a particle is moving along a circular path about a point as the centre then the motion of that particle is called circular or rotational motion.
The moment of linear momentum of a particle or a rigid body about an axis is known as angular momentum about that axis.
$\overrightarrow \tau = \overrightarrow I\,\overrightarrow \alpha $ , here $\overrightarrow \tau = $ torque acting on the particle of rigid body, I= moment of inertia and $\alpha $= angular acceleration.
If the torque acting on a body becomes 0 that is $\overrightarrow \tau = 0$ then the rate of change of angular momentum $(\overrightarrow L )$ also becomes zero. $\dfrac{{d\overrightarrow L }}{{dt}} = 0 \Rightarrow \overrightarrow L = cons\tan t$
If no external torque acts on a system of particles then, the total angular momentum of the system remains conserved.
Initially when the performance begins the person stretches the hand outward from the body to gain the required angular momentum. The angular momentum of the body is $\overleftrightarrow I\overrightarrow \omega $ and here $\,\overrightarrow \omega $ is the angular velocity.
Let the moment of inertia and angular velocity at the start of the spinning is ${I_1}$ and ${\omega _1}$
When the acrobat curls his body and brings his hands and legs as close to his body then his moment of inertia changes and consequently his angular velocity. The moment of inertia decreases because the various parts of his body come closer to the axis of rotation.
Let the moment of inertia and angular velocity after the acrobat curls his body of the spinning is ${I_2}$ and ${\omega _2}$
${I_1} > {I_2}$
No external torque acts on the system so angular momentum remains constant.
$\overrightarrow L = \overrightarrow I\,\,\overrightarrow \omega $
So $\overrightarrow I\overrightarrow \omega = $ constant.
Therefore, I decrease $\overrightarrow \omega $ and the acrobat will rotate with much higher velocity to conserve angular momentum.
Hence the principle involved in the performance of a spinning chair circus acrobat is conservation of angular momentum.Hence option (A) is correct.
Note
One may think why conservation of linear momentum is not applicable. When the net external force is zero then the conservation of linear momentum is valid. But here this is circular motion so there exists torque to create angular momentum. So, option “B” will not be correct. During this performance the conservation of angular momentum plays a huge role. So, option “A” is correct.
Firstly, we will write down the formula of torque in terms of moment of inertia and angular acceleration. We know that the rate of change of angular momentum equals the torque. Here the external torque is zero. So, we know that the angular momentum remains conserved. So, we will calculate that when an acrobat curls his body and brings his hands and legs as close to his body then his moment of inertia decreases and consequently his angular velocity increases. So, we can say easily that the principle involved is conservation of angular momentum.
Complete step by step answer
If a particle is moving along a circular path about a point as the centre then the motion of that particle is called circular or rotational motion.
The moment of linear momentum of a particle or a rigid body about an axis is known as angular momentum about that axis.
$\overrightarrow \tau = \overrightarrow I\,\overrightarrow \alpha $ , here $\overrightarrow \tau = $ torque acting on the particle of rigid body, I= moment of inertia and $\alpha $= angular acceleration.
If the torque acting on a body becomes 0 that is $\overrightarrow \tau = 0$ then the rate of change of angular momentum $(\overrightarrow L )$ also becomes zero. $\dfrac{{d\overrightarrow L }}{{dt}} = 0 \Rightarrow \overrightarrow L = cons\tan t$
If no external torque acts on a system of particles then, the total angular momentum of the system remains conserved.
Initially when the performance begins the person stretches the hand outward from the body to gain the required angular momentum. The angular momentum of the body is $\overleftrightarrow I\overrightarrow \omega $ and here $\,\overrightarrow \omega $ is the angular velocity.
Let the moment of inertia and angular velocity at the start of the spinning is ${I_1}$ and ${\omega _1}$
When the acrobat curls his body and brings his hands and legs as close to his body then his moment of inertia changes and consequently his angular velocity. The moment of inertia decreases because the various parts of his body come closer to the axis of rotation.
Let the moment of inertia and angular velocity after the acrobat curls his body of the spinning is ${I_2}$ and ${\omega _2}$
${I_1} > {I_2}$
No external torque acts on the system so angular momentum remains constant.
$\overrightarrow L = \overrightarrow I\,\,\overrightarrow \omega $
So $\overrightarrow I\overrightarrow \omega = $ constant.
Therefore, I decrease $\overrightarrow \omega $ and the acrobat will rotate with much higher velocity to conserve angular momentum.
Hence the principle involved in the performance of a spinning chair circus acrobat is conservation of angular momentum.Hence option (A) is correct.
Note
One may think why conservation of linear momentum is not applicable. When the net external force is zero then the conservation of linear momentum is valid. But here this is circular motion so there exists torque to create angular momentum. So, option “B” will not be correct. During this performance the conservation of angular momentum plays a huge role. So, option “A” is correct.
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