Physics Help!! Center of mass?

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Can you help me with this, I would owe you big.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?here is some additional informationImagine throwing a rock upward and away from you. With negligible air resistance, the rock will follow a parabolic path before hitting the ground. Now imagine throwing a stick (or any other extended object). The stick will tend to rotate as it travels through the air, and the motion of each point of the stick (taken individually) will be fairly complex. However, there will be one point that will follow a simple parabolic path: the point about which the stick rotates. No matter 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 then 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, it turns out, exists for every rigid object or system of massive particles. It is called the center of mass.To calculate the center of mass for a system of massive point particles that have coordinates (x_i, y_i) and masses m_i, the following equations are used:x_{rm cm}=frac{m_1x_1+m_2x_2+m_3x_3+cdots}{m_1+m_2+m_3+cdots},y_{rm cm}=frac{m_1y_1+m_2y_2+m_3y_3+cdots}{m_1+m_2+m_3+cdots}.In this problem, you will practice locating the center of mass for various systems of point particles.

Everardo Feil · 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 locatedThis 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 mmass 1 * distance from fulcrum = mass 2 * its distance from the fulcrumEq. #1.. m1 * d1 = m2 * d2Since they are sitting 10 m apartd1 + d2 = 10 Eq. #2.. d2 = 10 – d1Substitute into Eq. #1m1 * d1 = m2 * (10 – d1)divide both side by m1d1 = m2/m1 * (10 – d1)distribute right side d1 = 10 * m2/m1 – (m2/m1* d1)add (m2/m1* d1) to both sidesd1 + (m2/m1* d1) = 10 * m2/m1 factor d1 out of left sided1 (1 + m2/m1) = 10 * m2/m1divide 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 worksm1 = 30 kg m2 = 15 kgd1 = 10 * m2/m1 ÷ (1 + m2/m1)m2/m1 = 0.5d1 = 10 * 0.5 ÷ (1 + 0.5)di = 5 ÷ 1.5d1 = 3 ⅓ d2 = 10 – 3 ⅓ d2 = 6 ⅔check in torque equation #1Eq. #1.. m1 * d1 = m2 * d2Eq. #1.. 30 * 3 ⅓ = 15 * 6 ⅔100 = 100Solutiond1 = 10 * m2/m1 ÷ (1 + m2/m1)d2 = 10 – d1Let’s see if this can be simplified(1 + m2/m1) = (m1/m1 + m2/m1) = (m1 + m2)/m1d1 = 10 * m2/m1 ÷ (m1 + m2)/m1Let’s invert the second fraction and multiplyd1 = 10 * m2/m1 * m1/(m1 + m2)Look to the left and right of the multiply sign (*)/m1 * m1 = 1Better solution with all the stepsd1 = 10 * [m2 /(m1 + m2)]I am now retired from 23 years of teaching chemistry and physics and enjoy helping people understand math, chemistry, and physics.If you would like me to answer your questions in the future, request that I be one of your contacts. Then when you post a question, I will receive it via email. Otherwise I just search randomly for a question to answer.Thanks for the opportunity to be of help

Everardo Feil · 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

Everardo Feil · Answer

Let x be the distance of the centre of mass from the mass m1Then, by taking moments about the centre of massm1 . x = m2 ( 10 – x )(m1 . x) + ( m2 . x) = 10 . m2x ( m1 + m2 ) = 10 . m2x = 10 . m2 / ( m1 + m2 )

Mr. Durward Cummerata

Physics Help!! Center of mass?

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