9.1 Center of Mass
9.1.1 The Center of Mass
\overrightarrow r_{CM} = \frac {\sum m \overrightarrow r}{M_{total}}
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:
\overrightarrow p_{sys} = M_{total} \overrightarrow v_{CM}
* If the speeds are small compared to the speed of light
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