Question

A 1.00-m–long straight rod is manufactured to have a smoothly increasing linear mass density, so that at one end, it is 20.0 g/cm, and at the other end, it is 100 g/cm. (a) Assuming you place the low-mass-density end at x = 0, write the equation for the linear mass density of the rod as a function of x. (b) Use your answer from part (a) to determine the total mass of the rod. (c) Use your answers from parts (a) and (b) to find the location of the center of mass of the rod.

Answer 1

a) ρL(x)= 20+4/5*x

b) the total mass of the rod is M=6000gr = 6 Kg

c) the center of mass of the rod is X= 61.11 cm

Step-by-step explanation:

The linear mass density ρL is defined as

ρL(x)= dm(x)/dx , where m(x) is the mass of the rod at x (d represents the derivative operator)

a) since the linear density is a linear function:

ρL(x)= a+b*x

at x= 0 → ρL(x)=a =20 g/cm

at x=1 m=100 cm → ρL(x)=ρL(x)= 20 g/cm +b* (100 cm) = 100 g/cm → b= (100 g/cm - 20 g/cm)/100 cm = 4/5 g/cm²

ρL(x)= 20+4/5*x

b) considering that ρL(x)= dm(x)/dx → dm(x) = ρL(x) dx → m(x) = ∫ρL(x) dx + C

m(x) = ∫ρL(x) dx + C = ∫ (20+4/5*x) dx + C = 20x +4/5 (x²/2) + C = 2/5 x² + 20 x + C

now

m(0) = 0 → m(0) = 2/5 (0) + 20 (0) + C = C = 0

therefore

m(x) = 2/5 x² + 20 x

since the rod has 1m=100 cm long

M = m(100 cm) = 2/5 gr/cm² (100cm)² + 20 gr/cm (100cm) = 6000 gr = 6 Kg

c) the center of mass of the rod X is

X = (∫x dm) /M (evaluated between integration limits 0 and 100 cm of length)

but dm(x) = ρL(x) dx

X = (∫x ρL(x) dx ) /M = (1/M)*(∫ x*(20+4/5*x) dx ) = (1/M)*(∫ (20*x+4/5*x²) dx ) =

(1/M) ( 20(x²/2)+ 4/5*(x³/3) ) = (1/6000) ( 20*(100²/2) )+ 4/5*(100³/3) - 0 ) = 61.11 cm

therefore the center of mass X of the rod is X=61.11 cm