Question

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.

Answer 1

Answer:

Part (a)

$x_{cm}=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

Part (b)

x = 4.48 cm

Part (c)

$x_{cm}=\frac{-m_{1}d_{2}+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

Given that

First bead

mass , m₁ = 23g

d₁ = 1.1 cm

Second bead

mass, m₂ = 15 g

d₂ = 1.9 cm

Third bead

mass, m₃ = 58 g

d₃ = 3.2 cm

Part (a)

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

$x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

$x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}\\\\x_{cm}=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

Part (b)

the center of mass is

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

Part (c)

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

$x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

$x_{cm}=\frac{m_{1}-d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2} +m_{3}}\\\\x_{cm}=\frac{-m_{1}d_{2}+m_{3}d_{3} }{m_{1}+m_{2} +m_{3}}$

Part (d)

the center of mass is

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

Answer 2

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$