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2 years ago

Find the center of mass of a one-meter long rod, made of 50 cm of iron (density 8gcm3) and 50 cm of aluminum (density 2.7gcm3).

1 answer:

4 0

The center of the mass of a one-meter long rod, made of 50 cm of iron (density 8gcm3) and 50 cm of aluminum (density 2.7gcm3) is 37.62(cm)

<h3>Calculations and Parameters</h3>

λ Fe =8g/cm,λ Al =2.7g/cm

There is a metal rod 1 meter long, with the first 50cm being iron and the other 50cm aluminum. the problem is solved in a one-dimensional case meaning it's a thin rod.

Next, we would solve by definition of the center of mass

We would also note that the masses of iron and aluminum are just the product of their length and linear density.

$X_C_M$ = $\frac{8(g/cm). \frac{2500(cm^2)}{2} + 2.7 (g/cm). \frac{7500(cm^2)}{2} }{8(g/cm).50(cm) + 2.7(g/cm). 50 (cm)}$

= 37.62(cm)

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