Question

Find the center of mass (in cm) of a one-meter long rod, made of 50 cm of brass (density 8.44 g/cm3) and 50 cm of lead (density 11.3 g/cm3). (Assume the origin is at the midpoint of the rod, with the positive direction towards the part of the rod made of lead. Indicate the direction with the sign of your answer.)

Answer 1

Answer:

$\bar r = +3.622\,cm$

Explanation:

Let consider that both rods have uniform densities. The location of the center of mass is given by the following formula:

$\bar r = \frac{\rho_{brass}\cdot r_{brass}+\rho_{lead}\cdot r_{lead}}{\rho_{brass}+\rho_{lead}}$

$\bar r = \frac{(8.44\,\frac{g}{cm^{3}} )\cdot (-25\,cm)+(11.3\,\frac{g}{cm^{3}} )\cdot (+25\,cm)}{8.44\,\frac{g}{cm^{3}}+11.3\,\frac{g}{cm^{3}}}$

$\bar r = +3.622\,cm$

Answer 2

Answer:

Xcm=3.62cm

Explanation:

We can assume that the rods have the same radius (to compute the volume), an also we can assume that the center of mass of each part of the rod (for brasss and lead) is at the center of each material, that is, x1=-25cm, x2=25cm

If we take the origin at the center of the rod we have

$X_{CM}=\frac{M_1x_1+M_2x_2}{M_1+M_2}$

M1 (brass) and M2 (lead) are

$M_1=V*\rho_1=(\pi r^250cm)(8.44\frac{g}{cm})=422\pi r^2g\\M_2=V*\rho_2=(\pi r^250cm)(11.3\frac{g}{cm})=565\pi r^2g$

Hence,

$X_{CM}=\frac{(422\pi r^2g)(-25cm)+(565\pi r^2g)(25cm)}{422\pi r^2g+565\pi r^2g}\\X_{CM}=3.62cm$

hope this helps!!