The moment of the red box is the product of \(m_1\) and \(x_1\), and the moment of the blue box is product of \(m_2\) and \(x_2\). The seesaw is balanced with the fulcrum shifted to the left of the centre of the seesaw, at x = 0, the "origin".
The centre of mass of the "system" along the x-axis is where the fulcrum is. Mathematically it can be represented as:
\[\bar{x}=\frac{m_1\times x_1 +m_2\times x_2 }{m_1+m_2}\]
where the capital M is the mass of the total system.
It is the sum of the moment of each "element" divided by the total mass of the system. This makes intuitive sense.
Note this analysis is in one dimension. If we extend the analysis to two dimensions. As more than one dimension is involved one needs to be aware of the "axis" around which the moment will act. However, the mathematics are basically the same - except we need to do the calculation along the x and y-axes.
\[\bar{x}=\frac{\sum_{i=1}^{n}{m_i\times x_i}}{\sum_{i=1}^{n}{m_i}}\]
\[M_x=\sum_{i=1}^nm_iy_i\]
and
\[M_y=\sum_{i=1}^nm_ix_i\]
Also, the coordinates of the center of mass \((\bar{x},\bar{y})\) of the system are
\[\bar{x}=\dfrac{M_y}{m}\]
\[\bar{y}=\dfrac{M_x}{m}\]
\[M_x=\lim_{Δm→0}\sum_{i=1}^nΔm_iy_i =\int_{0}^{n}ydm\]
\[\M_y=\lim_{Δm→0}\sum_{i=1}^nΔm_ix_i =\int_{0}^{n}xdm]
Similarly, the moment of an element around the y-axis is the product of its mass and its distance from the axis, ie x.
Centre of Mass
The x-coordinate of the centre of mass is the sum of the moments about the y-axis, divided by the total mass of the system
\[\bar{x}=\frac{m_1\times x_1 +m_2\times x_2 }{m_1+m_2}=\frac{1}{M_T}\sum_{i=0}^{n}{m_i\times x_i}\]
Simiarly, the y-coordinate of the centre of mass is the moment of the whole system around the x-axis divided by the total mass of the system.
\[\bar{y}=\frac{m_1\times y_1 +m_2\times y_2 }{m_1+m_2}=\frac{1}{M_T}\sum_{i=0}^{n}{m_i\times y_i}\]


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