>[!summary] Angular momentum is used to describe how much force an object has when rotating. > This type of momentum is not always conserved nor is an intrinsic property > A object can be moving linear but still have angular momentum > > **Key equations:** **In** general this equation is true: $L = \vec{r} \times p$ > Special case where a body is rotating symmetrically: $L = I\omega$ >[!info]+ Read Time **⏱ 5 mins** # General Principle Angular momentum is used to describe an how much force an object has when rotating. It's similar to [[Linear Momentum]] in that it **can be** a conversed quantity but is **not an intrinsic property nor is always conserved!** >[!warning] Angular momentum does not **ALWAYS** mean spinning in a circle Although most situations of angular momentum is a object spinning in a circle hence having angular momentum. You can also have angular momentum if a object has [[Linear Momentum]] and has a side-way component (not moving straight) ![[ang_1.png]] [^1] >[!note] Explanation Example of angular momentum Angular momentum determine how much a object has rotational motion and how hard it will be to stop that. In our definition we use [[Linear Momentum]] crossed with the r vector. We do this because the r vector will determine the radius a object would rotate around some origin. We can choose any arbitrary origin point. General equation: $L = \vec{r} \times p$ ![[ang_4.png]] [^2] >[!note] Explanation Example of angular momentum # Special Case For Rotating Rigid Bodies >[!warning] Assumption For this special case well assume that our object is rotating symmetrically and the equation derived is for this special case (assume origin in the middle) > $v = \omega \times r$ from [[Rotational Kinematics]] not assuming $\omega$ and r are always perp from each other. ![[ang_2.png]] [^1] >[!note] Explanation A rotating rigid body rotating symmetrically. $\begin{array}{c} L = r \times p \\ L = r \times (mv) \\ L = mr \times (\omega \times r) \\ L = mr^2 \omega \\ L = I\omega \\ \end{array}$ # Conservation Important Example Angular momentum is conversed in special cases where we allow the center-mass (CM) to rotate systmmertaly. >[!warning] Assumptions In order to solve for $v, v', \omega$ well need to assume converation of linear and angular momentum from [[Equilibrium]] and [[Linear Momentum]] along with conservation of energy from [[Work, Energy & Power]] (The example is not in equilibrium, but uses equations from it ) ![[ang_3.png]] [^1] >[!note] Explanation Example of a stick and puck collision The puck reflects while the stick rotates in opposite direction of the reflected puck > **Important:** The v' velocity is technically to the right! >[!warning] Conversation of Linear Momentum If we isolate the case where $\sum F = 0$ so $\Delta p = 0$ $\begin{array}{c} p_i = mv +0 \\ p_f = mv' +Mv_{cm} \\ \\ \Delta p = 0\\ mv = mv' +Mv_{cm} \quad(1) \end{array}$ >[!warning] Conversation of Angular Momentum If we isolate the case where $\sum \tau = 0$ so $\Delta L = 0$ $\begin{array}{c} L_i = r\times p = hmv \quad \text{Assume into the page is pos} \\ L_f = r\times p +I\omega \quad \text{Assume into the page again}\\ L_f = hmv' + I\omega \\ \\ \Delta L = 0 \\ hmv = hmv' + I\omega_{cm} \quad (2)\\ \end{array}$ >[!warning] Energy Conversation Using conversation of energy from [[Work, Energy & Power]] well assume our system is made up of pure kinetic energy from linear and angular motion. (From [[Rotational Dynamics]] and idea from [[Rolling Motion With No Slipping]]) $\begin{array}{c} K_i = \frac{1}{2}mv^2 + 0 \\ K_f = \frac{1}{2}mv'^2 + \frac{1}{2}Mv_{cm}' + \frac{1}{2}I\omega ^2 \\ mv^2 = mv'^2_{cm} +\underbrace{\frac{1}{2}Mv_{cm}'}_{Stick \space Translaton} + \underbrace{I\omega ^2_{cm}}_{Stick \space Rotational} \quad (3)\\ \end{array} $ Using the three equations well solve for $v, v'$ and $\omega$ $\begin{array}{c} \text{Rewrite 1,2,3} \\ (1) \quad v-v' = \frac{M}{m}v_{cm} \\ (2) \quad v-v' = \frac{I}{hm} \omega_{cm} \\ (3) \quad (v-v')(v+v) = \frac{M}{m} v_{cm}^2 + \frac{I}{m} \omega_{cm}^2 \\ 1=> 2 \\ \frac{I}{hm} \omega_{cm} = \frac{M}{m}v_{cm} \\ \omega_{cm} = \frac{hM}{I} v_{cm} \\ \\ 2 = > 3 \\ (\frac{I}{mh}\omega_{cm})(v+v') = \frac{M}{m} v_{cm}^2 + \frac{I}{m} \omega_{cm}^2 \\ Note : \quad \omega_{cm} = \frac{hM}{I} v_{cm} \\ (\frac{M}{m}v_{cm})(v+v') = \frac{M}{m} v_{cm}^2 + \frac{M^2 h^2}{m}v_{cm} ^2 \\ (v +v') = v_{cm} + \frac{Mh^2}{I}v_{cm} \\ (v+v') = (1+\frac{Mh^2}{I}) v_{cm} \end{array}$ >[!bug] Warning We know the following things so far: >$\begin{array}{c} (v+v') = (1+\frac{Mh^2}{I}v_{cm}) \\ (v-v') = \frac{M}{m}v_{cm}\\\\ \text{Then we will create two new equations from this defniation} \\ (v+v') +(v-v') = 2v \\ (v+v') - (v-v') = 2v' \\ \end{array}$ $\begin{array}{c} (1+\frac{Mh^2}{I}v_{cm}) + \frac{M}{m}v_{cm} = 2v \\ v_{cm}(1+\frac{Mh^2}{I} + \frac{M}{m}) = 2v \\ v_{cm} = \frac{2}{(1+\frac{Mh^2}{I} + \frac{M}{m})} v \\ \\ (1+\frac{Mh^2}{I}v_{cm}) - \frac{M}{m}v_{cm} = 2v' \\ v_{cm}(1+\frac{Mh^2}{I} - \frac{M}{m} ) = 2v' \\ v' = \frac{v_{cm}(1+\frac{Mh^2}{I} - \frac{M}{m} )}{2} \\ \text{We should sub in vcm in best practices} \end{array}$ [^1]: Taken from R. Epp Lecture notes. [^2]: Taken from https://tikz.net/dynamics_angular_momentum/ from Izaak Neutelings (October 2020) --- > 💡 Found this concept helpful? [Star Math & Matter on GitHub](https://github.com/rajeevphysics/Obsidan-MathMatter) to support more intuitive science breakdowns like this. ---