Find the center of mass (in cm) of a one-meter long rod, made of 50 cm of copper (density 8.92 g/cm3) and 50 cm of iron (density 7.86 g/cm3). (Assume the origin is at the midpoint of the rod, with the positive direction towards the part of the rod made of iron. Indicate the direction with the sign of your answer.)

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Answer

given,

copper rod length = 50 cm

density of the copper = 8.92 g/cm³

iron rod length = 50 cm

density of iron = 7.86 g/cm³

mass of iron = density × volume

               m_i = 7.86 × A × l/2

               m_c = 8.96 × A × l/2

taking  the intersection of copper and iron rod be starting point cooper side is taken as positive side and iron side length is taken to be  -ve side.

center of mass

 = [tex]\dfrac{m_i\times \dfrac{-l}{4}+m_c\times \dfrac{l}{4}}{m_i+m_c}[/tex]

 = [tex]\dfrac{7.86\times A \times \dfrac{l}{2}\times \dfrac{-l}{4}+8.96\times A \times \dfrac{l}{2}\times \dfrac{l}{4}}{7.86\times A \times \dfrac{l}{2}+8.96\times A \times \dfrac{l}{2}}[/tex]

 = [tex]\dfrac{7.86\times \dfrac{-l}{4}+8.92\times \dfrac{l}{4}}{7.86+8.92}[/tex]

 = [tex]\dfrac{1.06\dfrac{l}{4}}{16.78}[/tex]

 = 0.015793 m

 = 1.579 m (+ve)

center of mass shift to cooper because cooper is heavy.

The center of mass is at -1.58 cm, that is towards the left of the origin, or towards the copper end.

Given information:

Let the length of the rod is 2l = 100cm = 1m, then

copper rod length l = 50 cm

density of the copper = 8.92 g/cm³

iron rod length l = 50 cm

density of iron = 7.86 g/cm³

Let the cross-sectional area of the rod be A, so

mass of iron = density × volume

[tex]m_i = 7.86 \times A \times l[/tex]

mass of copper

[tex]m_c = 8.92 \times A \times l[/tex]

Center of mass :

According to the question, the origin lies at the midpoint of the rod, that is the position of the joint between iron and copper.

let us assume that the copper rod is at the left of the origin and the iron rod is at the right of the origin, and both the rods are uniform, so the center of mass of the individual rods be at their midpoint, that is [tex]\frac{l}{2}[/tex], so the center of mass of the combined rod is :

[tex]cm = \frac{m_c\times(-l/2)+m_i\times(l/2)}{m_c+m_i}\\\\cm= \frac{ 8.92 \times A \times l\times(-l/2)+7.86 \times A \times l\times(l/2)}{8.92 \times A \times l+7.86 \times A \times l} \\\\cm=\frac{-0.53\times l}{16.78}=\frac{-0.53\times50}{16.78}\\\\cm= -1.58\;cm[/tex]

The negative sign indicates that the center of mass is towards the left.

Learn more about the center of mass:

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