11-07-25 Angular Momentum

Total Angular Momentum

• An object that is spinning around its center of mass while the center of mass while the center of mass is rotating around an external axis will have an angular momentum with contributions from both the translational and the rotational motion

• Thankfully, the total angular momentum of the system is simply the vector sum of the translational angular momentum and the rotational angular momentum:

$\overrightarrow L_{total,A} = \overrightarrow L_{trans,A} + \overrightarrow L_{rot}$

• The Angular Momentum Principle is very similar to the Momentum Principle as it can be written as (change in momentum) = (appropriate type of impulse):

$\Delta \overrightarrow L_{tot,A} = \overrightarrow \tau_{net,A} \Delta t$

Q10.x

A small rock passes a massive star. When the rock is a distance $3.9 \cdot 10^{13}$ m from the center of the star, the magnitude of its velocity is $1.1 \cdot 10^4$m/s, and the angle $\alpha$ is $117$ degrees. At a later time, when the rock is a distance $1.09 \cdot 10^{13}m$ along the x-axis from the center of the star, it is heading in the -y direction. There are no other massive objects nearby. What is the magnitude of the rock’s velocity at the later time?

$A)$ $1.10 \cdot 10^4$ $m/s$

$B)$ $1.79 \cdot 10^4$  $m/s$

$C)$ $3.51 \cdot 10^4$ $m/s$

$D)$ $3.94 \cdot 10^4$ $m/s$

$E)$ $2.74 \cdot 10^4$ $m/s$

Gather)

$|\overrightarrow r_1 | = 3.9 × 10^{13}$ $m$

$|\overrightarrow v_1| = 1.1 × 10^4$ $m/s$

$\theta_1 = \alpha = 117 \degree$

$|\overrightarrow r_2| = 1.09 × 10^{13}$ $m$

$|\overrightarrow v_2 | =$ $??$

$\theta_2 = 90\degree$

Analyze)

 $|\overrightarrow \tau_{net,A}| = |\overrightarrow r_A||\overrightarrow F_g| \sin (180\degree) = 0$

$\Delta \overrightarrow L_{tot,A}= 0 \cdot \Delta t \Rightarrow \overrightarrow L_{tot,A,f}=\overrightarrow L_{tot,A,i}$

$\Rightarrow |\overrightarrow L_{tot,A,f}| = |\overrightarrow L_{tot,A,i}|$

\Rightarrow |\overrightarrow r_1| |m\overrightarrow v_1|\sin(\theta_1) = |\overrightarrow r_2| |m\overrightarrow v_2|\sin(\theta_2)

$\Rightarrow |\overrightarrow v_2| = \frac {|\overrightarrow r_1||\overrightarrow v_1|\sin(\theta_1}{|\overrightarrow r_2|\sin(\theta_2)} = \frac {(3.9 × 10^{13}m)(1.1 × 10^4m)\sin(117\degree)}{(1.09 × 10^{13}m)\sin(90\degree} = 3.51 × 10^4 \frac ms$

The answer is C

Two children are attempting to balance themselves on a seesaw that has a mass of $10kg$ and a length of $4m$. The left edge of the seesaw is $1.1m$ away from its fulcrum. The child on the left has a mass of $60kg$ and sits on the left edge whereas the child on the right has a mass of $35kg$ and is moving to try to find balance. How far from the fulcrum should the child on the right be in order to balance the seesaw? Note: Gravity should be modeled ass acting on an objects center of mass.

$A)$ $1.89 m$

$B)$ $1.63m$

$C)$ $1.56 m$

$D)$ $0.64m$

$E)$ $0.53m$

Gather)

$m_S = 10kg$

$l_S = 4m$

$|\overrightarrow r_L| = 1.1m$

$m_L = 60kg$

$m_R=35kg$

$|\overrightarrow r_R| =$ $??$

$\overrightarrow L_{seesaw} = 0$

$\Delta \overrightarrow L_{seesaw} = 0$

Analyze)

$\Delta\overrightarrow L_{tot,A} = \overrightarrow \tau_{net,A} \Delta t \Rightarrow \overrightarrow \tau_{net,A} = 0 \Rightarrow$

$\overrightarrow \tau_{gL,A} + \overrightarrow \tau_{gS,A} + \overrightarrow \tau_{gR,A} = 0$

• \overrightarrow \tau_{gL,A} = \overrightarrow r_L × \overrightarrow F_{gL} = |\overrightarrow r_L||\overrightarrow F_{gL}|\sin(90\degree)<0,0,+1> 

• \overrightarrow \tau_{gS,A} = \overrightarrow r_S × \overrightarrow F_{gS}=|\overrightarrow r_S||\overrightarrow F_{gS}|\sin(90\degree)<0,0,-1>

• \overrightarrow \tau_{gR,A} = \overrightarrow r_R × \overrightarrow F_{gR}=|\overrightarrow r_R||\overrightarrow F_{gR}|\sin(90\degree)<0,0,-1>

$z:$ |\overrightarrow r_L||\overrightarrow F_{gL}| - |\overrightarrow r_S||\overrightarrow F_{gS}| - |\overrightarrow r_R||\overrightarrow F_{gR}|=0

$\Rightarrow |\overrightarrow r_L| m_Lg - (\frac {l_s}2 - |\overrightarrow r_L|)m_Sg-|\overrightarrow r_r|m_Rg=0$