Question

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.)

Answer 1

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

 = $\dfrac{m_i\times \dfrac{-l}{4}+m_c\times \dfrac{l}{4}}{m_i+m_c}$

 = $\dfrac{7.86\times A \times \dfrac{l}{2}\times \dfrac{-l}{4}+8.96\times A \times \dfrac{l}{2}\times \dfrac{l}{4}}{7.86\times A \times \dfrac{l}{2}+8.96\times A \times \dfrac{l}{2}}$

 = $\dfrac{7.86\times \dfrac{-l}{4}+8.92\times \dfrac{l}{4}}{7.86+8.92}$

 = $\dfrac{1.06\dfrac{l}{4}}{16.78}$

 = 0.015793 m

 = 1.579 m (+ve)

center of mass shift to cooper because cooper is heavy.