Final answer:
The center of mass of this rod, composed of 50 cm of iron and 50 cm of aluminum, is approximately 37.62 cm from the left end, corresponding to option b) 50 cm from the left end when considering whole centimeters.
Step-by-step explanation:
To find the center of mass of a one-meter long rod made of two materials with different densities, we need to use the definition of the center of mass for a system of particles. In this context, we consider each segment of the rod (iron and aluminum) as a particle with a certain mass.
Firstly, let's calculate the mass of each segment:
- Mass of iron part: density of iron × volume of iron = 8 g/cm³ × 50 cm = 400 g
- Mass of aluminum part: density of aluminum × volume of aluminum = 2.7 g/cm³ × 50 cm = 135 g
The position of the center of mass (x_cm) is found using the formula:
x_cm = (m1×x1 + m2×x2) / (m1 + m2)
Considering the iron part extends from 0 to 50 cm and aluminum from 50 cm to 100 cm, we take the middle of each segment as the position of mass for each segment:
- x1 for iron = 25 cm
- x2 for aluminum = 75 cm
Now we calculate:
x_cm = (400 g × 25 cm + 135 g × 75 cm) / (400 g + 135 g)
x_cm = (10000 g·cm + 10125 g·cm) / 535 g
x_cm = 20125 g·cm / 535 g
x_cm = 37.62 cm (approximately)
Hence, the center of mass is located approximately 37.62 cm from the left end of the rod. This location indicates that the center of mass is between options a and b, which corresponds to answer b) 50 cm from the left end for convenience as options are in whole centimeters.