The same principles that hold with linear momentum conservation can be applied here with angular momentum conservation. The direction of *L * is given by the right hand rule. Simply wrap your fingers around and in the direction the object is spinning and your thumb tells you the direction the vector is pointing.

Angular momentum can not change unless an outside torque is applied to the object.

Recall that momentum is a vector quantity, thus the direction a spinning object is pointing can not change without an applied torque.

\begin{align*}L = I\omega\end{align*}

\begin{align*}\vec{L} = \vec{r} \times \vec{p} = r _\perp p = rp _\perp\end{align*}

The *angular momentum* of a spinning object can be found in two equivalent ways. Just like linear momentum, one way, shown in the first equation, is to multiply the moment of inertia, the rotational analog of mass, with the angular velocity. The other way is simply multiplying the linear momentum by the radius, as shown in the second equation.

\begin{align*}\tau = \frac{\Delta L}{\Delta{t}}\end{align*}

Just the same as linear momentum, the torque required to change the momentum L in t time (L/t) can be compared to the force required to change the momentum p in t time. (p/t) Torques produce a change in angular momentum with time.

### Interactive Simulation

### Review

- You have two coins; one is a standard U.S. quarter, and the other is a coin of equal mass and size, but with a hole cut out of the center.
- Which coin has a higher moment of inertia?
- Which coin would have the greater angular momentum if they are both spun at the same angular velocity?

- A star is rotating with a period of 10.0 days. It collapses with no loss in mass to a white dwarf with a radius of \begin{align*}.001\end{align*}
.001 of its original radius.- What is its initial angular velocity?
- What is its angular velocity after collapse?

- A merry-go-round consists of a uniform solid disc of \begin{align*}225 \;\mathrm{kg}\end{align*}
225kg and a radius of \begin{align*}6.0 \;\mathrm{m}\end{align*}6.0m . A single \begin{align*}80 \;\mathrm{kg}\end{align*}80kg person stands on the edge when it is coasting at \begin{align*}0.20\end{align*}0.20 revolutions per sec. How fast would the device be rotating after the person has walked \begin{align*}3.5 \;\mathrm{m}\end{align*}3.5m toward the center. (The moments of inertia of compound objects add.)

**Review (Answers)**

- a. Coin with the hole b. Coin with the hole
- a. \begin{align*}7.27 \times 10^{-6} \;\mathrm{rad/s}\end{align*}
7.27×10−6rad/s b. \begin{align*}7.27 \;\mathrm{rad/s}\end{align*}7.27rad/s - \begin{align*}0.30 \;\mathrm{rev/sec}\end{align*}
0.30rev/sec