Question

A circus acrobat while performing spins brings his limbs closer to the body, due to this angular momentum is __________ but kinetic energy is ____________.

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Solution

**Angular momentum**

- When a circus acrobat performs spins.
- When no external torque acts on an object, there is no change in angular momentum that will occur.
- The angular momentum of a system is conserved as long as there is no net external torque acting on the system.
- Here, since there is no external torque acting on the body,
**angular momentum is conserved.** - Angular momentum, $L=I\omega $ where $\omega $ is the angular velocity and $I$ is the moment of inertia.
- The moment of inertia reduces as he moves his limbs closer to the body as $I=M{R}^{2}$ and $I\propto {R}^{2}$ where $M$ is the mass of ring and $R$ is the radius of ring.
- Due to the law of conservation of angular momentum, in the equation for angular momentum, when moment of inertia decreases, angular velocity increases. Since the angular velocity increases, the body rotates faster.
- So, angular momentum remains conserved

**Kinetic energy**

- He brings his limbs closer to his body while spinning.
- The moment of inertia reduces as he moves his limbs closer to the body.

as $I=M{R}^{2}$ and $I\propto {R}^{2}$ where $M$ is the mass of ring and $R$ is the radius of ring. - Angular momentum, $L=I\omega $ where $\omega $ is the angular velocity and $I$ is the moment of inertia.
- Kinetic energy, $K=\frac{1}{2}I{\omega}^{2}$ where $\omega $ is the angular velocity and $I$ is the moment of inertia.
- From the equation of angular momentum, since the angular momentum is conserved, we know that when moment of inertia $I$ decreases, angular velocity $\omega $ increases.
- Here, in case of kinetic energy, $K\propto I{\omega}^{2}$
- So, for example, if $I$ is doubled, by the equation of angular momentum,

$L=I\omega \phantom{\rule{0ex}{0ex}}\omega =\frac{L}{I}$

The new angular momentum becomes,

$L=I\text{'}\omega \text{'}\phantom{\rule{0ex}{0ex}}L=\left(2I\right)\omega \text{'}$

Angular frequency,

$\omega \text{'}=\frac{L}{2I}\phantom{\rule{0ex}{0ex}}\omega \text{'}=\frac{1}{2}\omega $

Kinetic energy,

$K=\frac{1}{2}I{\omega}^{2}$

So, new kinetic energy becomes

$K\text{'}=\frac{1}{2}I\text{'}\omega {\text{'}}^{2}\phantom{\rule{0ex}{0ex}}K\text{'}=\frac{1}{2}\left(2I\right){\left(\frac{\omega}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}K\text{'}=\frac{1}{2}I{\omega}^{2}\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}K\text{'}=K\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}K\text{'}=\frac{K}{2}$

From this example, it is clear that when $I$ increases, kinetic energy decreases.

When the moment of inertia of the body decreases, rotational kinetic energy increases because work is done to reduce the moment of inertia of the body.

As a result, final kinetic energy will exceed initial kinetic energy.

**So the K.E. is not conserved.**

Hence, a circus acrobat, while performing spins brings his limbs closer to the body, due to this angular momentum, is ** conserved **but kinetic energy is

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