Angular Momentum (AQA A Level Physics): Revision Note

A horizontal rigid bar is pivoted at its centre so that it is free to rotate. A point particle of mass $3M$ is attached at one end of the bar and a container is attached at the other end, both are at a distance of $R$ from the central pivot. 

A point particle of mass $M$ moves with velocity $v$ at right angles towards the stationary rod as shown in the diagram.

The particle $M$ collides with the container which is at a distance of $R$ just before the collision from the centre of the point of rotation of the bar and stays within it as the system starts to rotate about the vertical axis with angular velocity $\omega$.

The mass of the rod and the container are negligible.

Write an expression for the angular momentum of the system about the vertical axis just after the collision, in terms of $M$, $R$ and $\omega$.

[2]

Answer:

• The whole system rotates with an angular velocity of $\omega$

• The rod and the container are massless, so only the angular momentum of the two masses $M$ and $3M$ needs to be considered

• Therefore, the angular momentum of the system after the collision is:

$L=\sum \left(I\omega \right)$ [1 mark]

$L=\left(M{R}^2\right)\omega +\left(3M{R}^2\right)\omega$

$L=4M{R}^2\omega$ [1 mark]

You should know that objects travelling in straight lines can have angular momentum - just make sure you understand that it all depends on the position of the object in relation to the axis of rotation being considered.

The diagram shows the different positions of a diver between jumping off a springboard and entering the water.

During their fall, the diver pulls their arms and legs into a tight tuck position while in the air and straightens them before entering the water.

Which row correctly describes the changes to the diver's moment of inertia and angular velocity as they bring their limbs closer to their body?

[1]

Answer:

• After the diver leaves the springboard, there is no longer a resultant torque acting on them

  • This means their angular momentum remains constant throughout the dive

• Due to the conservation of angular momentum:

$I_i{\omega }_i=I_f{\omega }_f=\mathrm{constant}$

• When the diver tucks their arms and legs in closer to their body, they decrease their moment of inertia

  • This eliminates options A & C

• Therefore, to conserve angular momentum, when the diver's moment of inertia decreases, their angular velocity must increase

• This is option B [1 mark]

A spherical star of mass $M$ and radius $R$ rotates about its axis. The star explodes, ejecting mass in space radially and symmetrically. The remaining star is left with a mass of $\frac{1}{10}M$ and a radius of $\frac{1}{50}R$.

Calculate the ratio of the star’s final angular velocity to its initial angular velocity.

The moment of inertia of a sphere is $\frac{2}{5}M{R}^2$.

[2]

Answer:

• Before the star explodes:

  • Initial moment of inertia, $I_i=\frac{2}{5}M{R}^2$

  • Initial angular velocity = ${\omega }_i$

• After the star explodes:

  • Final moment of inertia, $I_f=\frac{2}{5}\left(\frac{1}{10}M\right){\left(\frac{1}{50}R\right)}^2$

  • Final angular velocity = ${\omega }_f$

• From the conservation of angular momentum:

$I_i{\omega }_i=I_f{\omega }_f$

$\left(\frac{2}{5}M{R}^2\right){\omega }_i=\left(\frac{2}{5}\times \frac{1}{10}M\times {\left(\frac{1}{50}\right)}^2{R}^2\right){\omega }_f$ [1 mark]

$\left(\cancel{\frac{2}{5}M{R}^2}\right){\omega }_i=\left(\cancel{\frac{2}{5}M{R}^2}\right)\frac{1}{25000}{\omega }_f$

• Therefore, the ratio of the star’s final angular velocity to its initial angular velocity is:

$\frac{{\omega }_f}{{\omega }_i}=25000$ [1 mark]