Answer:
Part a)
Center of mass with respect to the left end is given as
[tex]r_{cm} = 5.63 cm[/tex]
Part b)
Center of mass with respect to middle bead is
[tex]r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}[/tex]
Part c)
Center of mass with respect to middle bead is
[tex]r_{cm} = 1.63 cm[/tex]
Explanation:
Part a)
As we know that the center of mass of the system of mass is given by the formula
[tex]r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}[/tex]
here we have
[tex]m_1 = 28 g[/tex]
[tex]m_2 = 11 g[/tex]
[tex]m_3 = 45 g[/tex]
[tex]r_1 = 1.5 cm[/tex]
[tex]r_2 = 1.5 + 2.5 = 4 cm[/tex]
[tex]r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm[/tex]
Now we have
[tex]r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}[/tex]
[tex]r_{cm} = 5.63 cm[/tex]
Part b)
As we know that the center of mass of the system of mass is given by the formula
[tex]r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}[/tex]
here we have
[tex]m_1 = 28 g[/tex]
[tex]m_2 = 11 g[/tex]
[tex]m_3 = 45 g[/tex]
[tex]r_1 = -d_2 = -2.5cm[/tex]
[tex]r_2 = 0[/tex]
[tex]r_3 = d_3 = 4.6 cm[/tex]
[tex]r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}[/tex]
Part c)
Now plug in the values in above formula
[tex]r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}[/tex]
[tex]r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}[/tex]
[tex]r_{cm} = 1.63 cm[/tex]