9.1 Center of Mass

9.1.1 The Center of Mass

Example 9.1.A | Two Rocks

A 10 kg rock is located at <3,0,0> m, and a $2 kg$ rock is located at <8,2,0> m. Find the center of mass of the two-rock system.

$r_1 =$ <3,0,0> m 

$m_1 =$ $10kg$

$r_2 =$ <8,2,0>m

$m_2 = 2 kg$

CM = \frac {(10 \cdot <3,0,0>) + (2 \cdot <8,2,0>)}{12} = \frac {<30,0,0> + <16,4,0>}{12} = \frac {<46,4,0>}{12}

$CM =$ <3.8\overline{33}, 0.\overline{33}, 0>

9.1.2 The Velocity and Momentum of the Center of Mass

$\overrightarrow v_{CM} = \frac {d}{dt} \overrightarrow r_{CM} = \frac {\sum m\overrightarrow v}{M_{total}}$

If the speeds are small compared to the speed of light ($\gamma = 1$), the quantity $\sum m\overrightarrow v$ is the total momentum of the system, so we can express the total momentum of a multiparticle system in terms of its center-of-mass velocity:

Checkpoint 9.1-C-01

A system consists of a 2 kg block moving with velocity < -3, 4, 0> m/s and a $5kg$ block moving with velocity <2,6,0> m/s. Calculate the velocity of the center of mass of the two-block system.

$A)$ <0.57, 5.43, 0> m/s

$B)$ <4.0, 38.0,0>m/s

$C)$ <10.0,30.0,0> m/s

$D)$ <-0.14, 1.43,9> m/s

$m_1 = 2kg$

$\overrightarrow v_1 =$ <-3,4,0> m/s

$m_2 = 5kg$

$\overrightarrow v_2 =$ <2,6,0> m/s

\overrightarrow v_{CM} = \sum \frac {m\overrightarrow v}{M_{total}} = \frac {(2kg\cdot <-3,4,0>m/s) + (5kg\cdot <2,6,0>m/s)} {7kg} 

\overrightarrow v_{CM} = \frac {(<-6,8,0>kg\cdot m/s)+(<10,30,0> kg\cdot m/s)}{7kg}

\overrightarrow v_{CM} = \frac {<4,38,0>kg\cdot m/s}{7kg}
$\overrightarrow v_{CM} =$ <0.571,5.429,0>m/s

The correct answer is A