The linear density of a rod,in g/m,is given by $\lambda = 40.0 + 30.0 x$ .The rod extends from the origin to x = 0.400 m.What is the location of the center of mass of the rod?
A)x = 0.213 m
B)x = 0.315 m
C)x = 0.384 m
D)x = 0.184 m
E)x = 0.208 m

Question

Quizplus · Accepted Answer

The center of mass $ x_{cm} $ is given by $ x_{cm} = \frac{\int x \lambda(x) \, dx}{\int \lambda(x) \, dx} $. Integrating from 0 to 0.4 m, we find $ x_{cm} = \frac{\int_0^{0.4} x (40 + 30x) \, dx}{\int_0^{0.4} (40 + 30x) \, dx} $. Solving these integrals gives $ x_{cm} = 0.213 \, \text{m} $.

The Linear Density of a Rod,in G/m,is Given By λ=40.0+30.0x\lambda = 40.0 + 30.0 xλ=40.0+30.0x

The Linear Density of a Rod,in G/m,is Given By $\lambda = 40.0 + 30.0 x$