Angular Momentum & Rolling - Physics Q and A

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Visualization of a spinning figure skater pulling her arms inward, surrounded by glowing vector lines representing conserved angular momentum.

If there is no external torque on a system, the universe absolutely forbids its angular momentum from changing.

In our linear mechanics units, we learned that momentum ( $p = mv$ ) is always conserved in a collision if there are no external forces. In the rotational world, the same rule applies: Angular Momentum ( $L$ ) is conserved as long as there is no net external torque.

1. Calculating Angular Momentum ( $L$ )

There are two ways to calculate Angular Momentum, depending on what the object is doing. If an object is a solid mass spinning on an axis (like a wheel or a figure skater), we use the rotational equivalent of $p = mv$ :

$L = I\omega$

However, if the object is a point particle moving in a straight line past a reference point (like an asteroid flying past a planet), it still has angular momentum relative to that point! We calculate this using the vector cross product:

$\vec{L} = \vec{r} \times \vec{p} = rmv \sin\theta$

Where $\vec{r}$ is the distance vector from the pivot to the particle, and $\vec{p}$ is the linear momentum.

2. Rolling Without Slipping

When a wheel rolls down a hill perfectly without skidding, it is doing two things simultaneously: it is translating (moving forward) and rotating (spinning). This means its total kinetic energy is split into two “bank accounts”.

$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Because the wheel isn’t slipping, the linear variables and rotational variables are locked together by the radius of the wheel ( $R$ ):

Velocity: $v = \omega R$
Acceleration: $a = \alpha R$

Concept First: Look at the simulator! The objects all start with the same gravitational potential energy ( $mgh$ ). But because the Hoop has the highest Moment of Inertia ( $I=MR^2$ ), it requires the most rotational energy to spin. That leaves less energy available for translation ( $v$ ), meaning the Hoop always loses the race to the Sphere ( $I=\frac{2}{5}MR^2$ )!

3. Quick AP Practice

📚 Unit C5.2 Mastery Challenge

1. A star rotating at an angular velocity $\omega_0$ collapses into a neutron star, decreasing its radius by a factor of 10. Assuming the star is a uniform solid sphere ( $I = \frac{2}{5}MR^2$ ) and no mass is lost, what is its new angular velocity in terms of $\omega_0$ ?

Check Answer

Use Conservation of Angular Momentum ( $L_i = L_f$ ).
$I_i \omega_i = I_f \omega_f$
$\left(\frac{2}{5}MR_i^2\right) \omega_0 = \left(\frac{2}{5}M(\frac{R_i}{10})^2\right) \omega_f$

Cancel out $\frac{2}{5}M$ :
$R_i^2 \omega_0 = \left(\frac{R_i^2}{100}\right) \omega_f$
$\omega_f = \mathbf{100 \omega_0}$ . (The star spins 100 times faster!)

2. A solid cylinder ( $I = \frac{1}{2}MR^2$ ) rolls without slipping down a ramp of height $h$ . Derive an expression for its final linear velocity $v$ at the bottom of the ramp.

Check Answer

Use Conservation of Energy: $E_i = E_f$ .
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$ .

Substitute $I = \frac{1}{2}MR^2$ and $\omega = \frac{v}{R}$ :
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v^2}{R^2}\right)$
$Mgh = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$ (The $R^2$ terms cancel)
$Mgh = \frac{3}{4}Mv^2 \Rightarrow gh = \frac{3}{4}v^2 \Rightarrow \mathbf{v = \sqrt{\frac{4gh}{3}}}$ .

1. Calculating Angular Momentum ()

2. Rolling Without Slipping

3. Quick AP Practice

📚 Unit C5.2 Mastery Challenge

🔗 Next Steps

1. Calculating Angular Momentum ( $L$ )