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 aluminum (density 2.7 g/cm3). (Assume the origin is at the midpoint of the rod, with the positive direction towards the part of the rod made of aluminum. Indicate the direction with the sign of your answer.)

Answer 1

Answer:

The center of mass is at + 37.11 cm

Solution:

As per the question:

Length of the rod, l = 1 m = 100 cm

Density of Brass, $\rho_{B} = 8.44\ g/cm^{3}$

Density of Aluminium, $\rho_{Al} = 2.7\ g/cm^{3}$

Distance from the mid point to each, r = 50 cm

Now,

The center of mass with the origin at mid point is given by the expression:

$X_{CoM} = \frac{\int_{0}^{50} \rho_{B} rdr + \int_{100}^{50} \rho_{Al} rdr}{M_{B}r + M_{Al}r}$

$X_{CoM} = \frac{\rho_{B}\frac{r^{2}}{2}]_{0}^{50} + \rho_{Al}\frac{r^{2}}{2}]_{100}^{50}}{M_{B}r + M_{Al}r}$

$X_{CoM} = \frac{8.44\frac{2500}{2}] + 2.7\frac{7500}{2}]{8.44\times 50 + 2.7\times 50}$

$X_{CoM} = + 37.11\ cm$