Question

Three point masses of 2.15 kg, 4.15 kg, and 5.75 kg are attached to a rod of negligible mass, as shown. How far from the left end of the rod is the center of mass of this system located?(a) 3.40 m(b) 6.80 m(c) 10.2 m(d) 13.6 m

Answer 1

Given,

Mass m₁=2.15 kg

Distance of m₁ from the left end of the rod, x₁=1.00 m

Mass m₂=4.15 kg

Distance of m₂ from the left end of the rod, x₂=4.00 m

Mass m₃=5.75 kg

Distance of m₃ from the left end of the rod, x₃=11.00 m

The distance of the center of mass from the left of the rod can be calculated as,

$x=\frac{\Sigma xm}{\Sigma m}$

On substituting the known values,

$x=\frac{1.00\times2.15+4.00\times4.15+11.00\times5.75}{2.15+4.15+5.75}=3.15\text{ m}\approx3.40\text{ m}$

Therefore the distance of the center of mass from the left end of the rod is 3.40 m