Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end of the rod. The second bead, of mass m2 = 15 g, is placed a distance d2 = 1.9 cm to the right of the first bead. The third bead, of mass m3 = 58 g, is placed a distance d3 = 3.2 cm to the right of the second bead. Assume an x-axis that points to the right.
a. Write a symbolic equation for the location of the center of mass of the three beads relative to the left end of the rod, in terms of the variables given in the problem statement.
b. Find the center of mass, in centimeters, relative to the left end of the rod.
c. Write a symbolic equation for the location of the center of mass of the three beads relative to the center of bead, in terms of the variables given in the problem statement.
d. Find the center of mass, in centimeters, relative to the middle bead.

[tex]x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}\\\\x_{cm}=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }[/tex]

Part (b)

the center of mass is

[tex]x_{cm}=\frac{23g*1.1cm+15g*(1.1cm+1.9cm)+58g(1.1cm+1.9cm+3.2cm) }{23g+15g+58g } \\\\x_{cm}= \frac{25.3+45+359.6}{96} \\\\x_{cm}=\frac{429.9}{96}\\\\x_{cm}=4.48cm[/tex]

Part (c)

the location of the center of mass of the three beads is

[tex]x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}[/tex]

[tex]x_{cm}=\frac{m_{1}-d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2} +m_{3}}\\\\x_{cm}=\frac{-m_{1}d_{2}+m_{3}d_{3} }{m_{1}+m_{2} +m_{3}}[/tex]

Part (d)

the center of mass is

[tex]x_{cm}=\frac{-(23g)(1.9cm)+(58g)(3.2cm) }{23g+15g+58g} \\\\ x_{cm} = \frac{-43.7+185.6}{96} \\\\ x_{cm}=\frac{141.9}{96} \\\\ x_{cm}=1.48cm[/tex]

oeerivona oeerivona · Answer 2 · 2021-11-01T19:05:09+01:00

The center of mass is the location relative to which the sum of the

distances of a distributed system of masses is zero.

The correct responses are;

[tex]a. \ x_{cm} = \dfrac{m_1 \times d_1 + m_2 \times (d_1 + d_2) + m_3 \times (d_1 + d_2 + d_3)}{m_1 + m_2 + m_3}[/tex]

b. The center of mass is 4.478125 cm from the left end.

[tex]c. \ The \ symbolic \ equation \ for \ the \ center \ of \ mass \ is \ x_{cmcb} = \dfrac{ m_3 \times d_3- m_1 \times d_2 }{m_1 + m_2 + m_3}[/tex]

d. [tex]x_{cmcb}[/tex] = 1.478125 cm

Reasons:

a. The center of mass of the three beads relative to the left of the rod, [tex]x_{cm}[/tex],

is given using the formula for center of mass as follows;

[tex]x_{cm} = \dfrac{m_1 \times d_1 + m_2 \times (d_1 + d_2) + m_3 \times (d_1 + d_2 + d_3)}{m_1 + m_2 + m_3}[/tex]

b. Given that we have;

m₁ = 23 g, m₂ = 15 g, m₃ = 58 g, d₁ = 1.1 cm, d₂ = 1.2, d₃ = 3.2 cm

Therefore;

[tex]x_{cm} = \dfrac{23 \times 1.1 + 15 \times (1.1 + 1.9) + 58 \times (1.1 + 1.9 +3.2)}{23 + 15 + 58} = \dfrac{1433}{320} = 4.478125[/tex]

[tex]x_{cm}[/tex] = 4.478125 cm from the left end

c. The center of mass of the beads with measurements taken relative to the

center bead, [tex]x_{cmcb}[/tex] is presented as follows;

[tex]x_{cmcb} = \dfrac{m_1 \times (-d_2) + m_2 \times (0) + m_3 \times d_3}{m_1 + m_2 + m_3} = \dfrac{ m_3 \times d_3- m_1 \times d_2 }{m_1 + m_2 + m_3}[/tex]

d. The center of mass relative to the middle bead is therefore;

[tex]x_{cmcb} = \dfrac{ 58 \times 3.2- 23 \times 1.9 }{23 + 15 + 58} = 1.478125[/tex]

[tex]x_{cmcb}[/tex] = 1.478125 cm

Learn more here:

https://brainly.com/question/21959097