Physics Help!! Center of mass?

Question

Can you help me with a physics problem? I would greatly appreciate it! 
Consider two particles with masses ( m_1 ) and ( m_2 ) (where ( m_1 < m_2 )) located 10 meters apart. I need to determine the location of the center of mass of this system. 
To provide some context, imagine throwing a rock upward and away from you. With negligible air resistance, the rock will follow a parabolic path before hitting the ground. Now, consider throwing a stick (or any other extended object). The stick tends to rotate as it travels through the air, and the motion of each point of the stick, taken individually, is rather complex. However, there exists one point that follows a simple parabolic trajectory: the point about which the stick rotates. Regardless of how the stick is thrown, this special point will always be located at the same position within the stick. 
The motion of the entire stick can be described as a combination of the translation of that single point (as if the entire mass of the stick were concentrated there) and the rotation of the stick about that point. Such a point exists for every rigid object or system of massive particles; it is known as the center of mass. 
To calculate the center of mass for a system of massive point particles with coordinates ((x_i, y_i)) and masses (m_i), the following equations are used:
[
x_{	ext{cm}} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + \cdots}{m_1 + m_2 + m_3 + \cdots}
]
[
y_{	ext{cm}} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + \cdots}{m_1 + m_2 + m_3 + \cdots}
]
In this problem, I would like to practice locating the center of mass for various systems of point particles. Thank you for your assistance!

Chelsie Ward · Accepted Answer

Two particles of masses m_1 and m_2 (m_1 < m_2) are located 10 meters apart. Where is the center of mass of the system located
This is a torque problem, similar to two children on a teeter-totter.
If the teeter-totter is adjusted so both children can have their feet off the ground and board does not move, the fulcrum is at the center of mass.
Mass 1 is on the left end of  the 10 m
mass 1 * distance from fulcrum = mass 2 * its distance from the fulcrum
Eq. #1.. m1 * d1 = m2 * d2
Since they are sitting 10 m apart
d1 + d2 = 10 
Eq. #2.. d2 = 10 – d1
Substitute into Eq. #1
m1 * d1 = m2 * (10 – d1)
divide both side by m1
d1 = m2/m1 * (10 – d1)
distribute right side 
d1 = 10 * m2/m1 – (m2/m1* d1)
add (m2/m1* d1) to both sides
d1 + (m2/m1* d1) = 10 * m2/m1 
factor d1 out of left side
d1 (1 + m2/m1) = 10 * m2/m1
divide both sides by (1 + m2)
d1 = 10 * m2/m1 ÷ (1 + m2/m1)
Let’s see make up a problem with masses see if this equation works
m1 = 30 kg 
m2 = 15 kg
d1 = 10 * m2/m1 ÷ (1 + m2/m1)
m2/m1 = 0.5
d1 = 10 * 0.5 ÷ (1 + 0.5)
di = 5 ÷ 1.5
d1 = 3 ⅓
 d2 = 10 – 3 ⅓
 d2 = 6 ⅔
check in torque equation #1
Eq. #1.. m1 * d1 = m2 * d2
Eq. #1.. 30 * 3 ⅓  = 15 * 6 ⅔
100 = 100
Solution
d1 = 10 * m2/m1 ÷ (1 + m2/m1)
d2 = 10 – d1
Let’s see if this can be simplified
(1 + m2/m1) = (m1/m1 + m2/m1) = (m1 + m2)/m1
d1 = 10 * m2/m1 ÷ (m1 + m2)/m1
Let’s invert the second fraction and multiply
d1 = 10 * m2/m1 * m1/(m1 + m2)
Look to the left and right of the multiply sign (*)
/m1 * m1 = 1
Better solution with all the steps
d1 = 10 * [m2 /(m1 + m2)]
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Anonymous · Answer

Two particles of masses m_1 and m_2 (m_1 < m_2) are located 10 meters apart. Where is the center of mass of the system located?
Let center of mass is d meters from m_1.
Then 
m_1x d  = (  m_2) x ( 10 – d)
d = 10 m_2 / (m_1 + m_2)………………………..ans

Anonymous · Answer

Let x be the distance of the centre of mass from the mass m1
Then, by taking moments about the centre of mass
m1 . x   =  m2 ( 10 – x )
(m1 . x)  + ( m2 . x)  = 10 . m2
x ( m1 + m2 )  = 10 . m2
x    =    10  .  m2 / ( m1 + m2 )

Cydney Leuschke

Physics Help!! Center of mass?

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