Question

Three beads are placed along a thin rod. The first bead, of mass m1 = 28 g, is placed a distance d1 = 1.5 cm from the left end of the rod. The second bead, of mass m2 = 11 g, is placed a distance d2 = 2.5 cm to the right of the first bead. The third bead, of mass m3 = 45 g, is placed a distance d3 = 4.6 cm to the right of the second bead. Assume an x-axis that points to the right.
(a) Find the center of mass, in centimeters, relative to the left end of the rod.
(b) Write a symbolic equation for the location of the center of mass of the three beads relative to the center bead, in terms of the variables given in the problem statement.
(c) Find the center of mass, in centimeters, relative to the middle bead.

Answer 1

Answer:

Part a)

Center of mass with respect to the left end is given as

$r_{cm} = 5.63 cm$

Part b)

Center of mass with respect to middle bead is

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Center of mass with respect to middle bead is

$r_{cm} = 1.63 cm$

Explanation:

Part a)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = 1.5 cm$

$r_2 = 1.5 + 2.5 = 4 cm$

$r_3 = 1.5 + 2.5 + 4.6 = 8.6 cm$

Now we have

$r_{cm} = \frac{28(1.5) + 11(4) + 45(8.6)}{28 + 11 + 45}$

$r_{cm} = 5.63 cm$

Part b)

As we know that the center of mass of the system of mass is given by the formula

$r_{cm} = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

here we have

$m_1 = 28 g$

$m_2 = 11 g$

$m_3 = 45 g$

$r_1 = -d_2 = -2.5cm$

$r_2 = 0$

$r_3 = d_3 = 4.6 cm$

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

Part c)

Now plug in the values in above formula

$r_{cm} = \frac{m_1(-d_2) + m_2(0) + m_3(d_3)}{m_1 + m_2 + m_3}$

$r_{cm} = \frac{28(-2.5) + m_2(0) + 45(4.6)}{28 + 11 + 45}$

$r_{cm} = 1.63 cm$