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183_notes:examples:walking_in_a_boat [2014/10/01 05:31] – pwirving | 183_notes:examples:walking_in_a_boat [2014/10/01 05:46] – pwirving | ||
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$\vec{v}{_{cm, | $\vec{v}{_{cm, | ||
- | $\vec{r}{_{cm}}$ is fixed $\longrightarrow$ both are zero. | + | Referring to the diagram above we can see that after the motion of the person the $\vec{r}{_{cm}}$ is the same and hence fixed $\longrightarrow$ |
+ | |||
+ | The center of mass of a system is the weighted average of the particles in that system. The center of mass of the system in question is the vector sum of each part of the systems mass by their location relative to the origin. | ||
Initially, | Initially, | ||
$\vec{r}{_{cm}} = \dfrac{1}{M_{tot}}\left(\sum_i m_i \vec{r}_i\right)$ in 1D, | $\vec{r}{_{cm}} = \dfrac{1}{M_{tot}}\left(\sum_i m_i \vec{r}_i\right)$ in 1D, | ||
+ | |||
+ | Therefore $x_{cm,i}$ initial is the mass of the boat by the location relative to the origin plus the mass of the person by their location relative to the origin divided by the total mass of the system. | ||
$x_{cm,i} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$ | $x_{cm,i} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$ | ||
- | In the final state, we don't know x, but we know that $x_{cm,f} = x_{cm,i}$ So we'll just use the unknown x. | + | In the final state, we don't know x, but we know that $x_{cm,f} = x_{cm,i}$ So we'll just use the unknown x and insert the distance into the distances for boat and person relative to the origin. |
$x_{cm,f} = \dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m}$ | $x_{cm,f} = \dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m}$ | ||
+ | |||
+ | As indicated the center of the mass of the system as not changed position therefore: | ||
$x_{cm,f} = x_{cm,i}$ | $x_{cm,f} = x_{cm,i}$ | ||
+ | |||
+ | So we can relate the equations for initial and final to each other: | ||
$\dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$ | $\dfrac {M(x+D+\dfrac{L}{2}) + m(x+D)}{M+m} = \dfrac {M(D+\dfrac{L}{2}) + m(D+L)}{M+m}$ | ||
- | Same denominator | + | Both of these equations have the same denominator |
${M(x+D+\dfrac{L}{2}) + m(x+D)} = {M(D+\dfrac{L}{2}) + m(D+L)}$ | ${M(x+D+\dfrac{L}{2}) + m(x+D)} = {M(D+\dfrac{L}{2}) + m(D+L)}$ | ||
- | Solve for x, | + | Multiply out to solve for x: |
- | $M_x + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$ | + | $Mx + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$ |
- | Cancel like terms | + | Cancel like terms we get: |
$Mx + mx = mL$ | $Mx + mx = mL$ |