Learn displacement formula in physics with momentum and center of mass. Practice velocity of center of mass and displacement of an object.
Case 1:
Case 2:
\(\vec{r} _{cm} = \dfrac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}\)
\(m_1 = m_1 \;\;\;\;\;\;\;\; x_1 = 0\)
\(m_2 = m_2 \;\;\;\;\;\;\; x_2 = d\)
\( = \dfrac{m_1 \times0 + m_2 \;d}{m_1 + m_2 }\, =\;\dfrac{m_2\; d}{m_1 + m_2 }\)
\(\vec{s}_{com} = 0\)
by applying
\(\vec{s}_{com} = 0\)
\(m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0\)
\(\vec{s}_{1} = x\) and \(\vec{s}_2 = - [x_1- (x_2 + x)]\) [from figure]
\(m_1 \; x - m_2 [x_1 - (x_2 + x)] = 0\)
\(m_1 \; x - m_2x_1 + m_2 (x_2 + x) = 0\)
\(m_1 \; x - m_2 x_1 + m_2 x_2 + m_2 x = 0\)
\((m_1 + m_2 ) x = m_ 2 (x_1 - x_2)\)
\(x = \dfrac{m_2 (x_1 - x_2 )}{m_1 + m_2 }\)
Thus, \(\vec{s}_{com} = 0\)
\(m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0\)
Here, \(\vec{s}_1 = -x, \,\vec{s}_2 = + (\ell - x)\)
\(m_1(-x) + m_2 [+ (\ell -x)] = 0\)
\(-m_1 \; x + m_2 (\ell-x) = 0\)
\(m_1\; x = m_2\; (\ell -x)\)
\(m_1\; x + m_2 \; x = m _2 \; \ell\)
\((m_1 + m_2) x = m _2 \; \ell\)
\(x = \dfrac{m_2 \; \ell }{m_1+ m_2}\)
\(\vec{x}_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}\)
Hence, displacement of center of mass along x-axis is zero.
\(\vec{s}_{com} = 0\)
\(m_1\vec{s}_1+ m_2 \vec{s}_2 = 0\) where \(\vec{s}_1 = - x, \, \vec{s}_2 = (\ell - x)\)
\(\Rightarrow - m_1x + m_2 [+ (\ell - x)] = 0\)
\(\Rightarrow - m_1x + m_2 (\ell - x) = 0\)
\(\Rightarrow m_1x = m_2 (\ell - x)\)
\(\Rightarrow m_1x = m_2\ell-m_2x\)
\(\Rightarrow (m_1 + m_2) x = m _2 \; \ell\)
\(\Rightarrow \; x = \left(\dfrac{m_2}{m_1 + m_2}\right) \ell\)
Let the center of mass be the origin.
\(\vec{s}_{com} = 0\)
\(m_1 \vec{s}_1 + m_2 \vec{s}_2 = 0\)
where \(\vec{s}_1 = + x, \,\vec{s}_2 = - (\ell - x)\)
So, \(m_1x+m_2 [- (\ell - x)] = 0\)
\(\Rightarrow m_1 x - m_2 \ell + m_2 x = 0\)
\(\Rightarrow (m_1 + m_2 )x = m_2 \ell \)
\(x = \dfrac{m_2 \; \ell }{m_1 + m_2}\)
(a) If \(\vec{v}\) is parallel to \(\vec{a}\) then straight line
(b) If \(\vec{v}\) is not parallel to \(\vec{a}\) then parabola.
Since \(m_1 = m_1 , \, m_2 = m_2\)
\(x_1 = -d , \, x _2 = d_2\)
\(x_{cm} = 0\)
\(o = \dfrac{m_1 (-d) + m_2 (d_2)}{m_1 + m_2 }\)
\(d_2 = \dfrac{m_1 \; d}{m_2}\)
(a) If \(\vec{v}\) is parallel to \(\vec{a}\) then straight line
(b) If \(\vec{v}\) is not parallel to \(\vec{a}\) then parabola.
Here \(m_1 = m_1 \, , \,m_2 = m_2\)
\(x_1 = R/2 \, , \, x_{cm} = R\)
\(R = \dfrac{m_1(R/2) + m_2 (x)}{m_1 + m_2 }\)
\(m_1 R + m_2 R = m_1 \dfrac{R}{2} + m_2 x\)
\(x = \dfrac{m_1 \dfrac{R}{2} + m_2 R}{m_2}\)