The density of a 5.0-m long rod can be described by the linear density function λ(x) = 145 g/m + 14.2x g/m2. One end of the rod is positioned at x = 0 and the other at x = 5 m.
a) Determine the total mass of the rod.
b) Determine the center-of-mass coordinate.
m=∫ρ•dx =∫(145+14.2x) •dx =
= ∫145•dx+∫14.2•x•dx =
=145•x + 14.2x²/2=
=145•5 + 14.2•25/2 =902.5 g.
Calculate the integral
∫ρ•x•dx =
=∫(145+14.2x) •x •dx =
= ∫145•x•dx+∫14.2•x²•dx =
=145•x²/2 + 14.2x³/3=
=145•25/2 + 14.2•125/3=2404.2 kg.
x(c/m/) =∫ρ•x•dx/∫ρ•dx =2404.2/902.5=2.66 m.
C.M. (2.66 m; 0)
Question: Shouldn't the 2404.2 be in grams and not Kg?
To solve this problem, we need to integrate the linear density function to find the total mass of the rod and calculate the center-of-mass coordinate.
a) To determine the total mass of the rod, we integrate the linear density function λ(x) over the length of the rod:
M = ∫ λ(x) dx
Given that the linear density function is λ(x) = 145 g/m + 14.2x g/m^2, we can substitute this into the integral:
M = ∫ (145 + 14.2x) dx
Integrating this function gives us:
M = 145x + 7.1x^2 ] from x = 0 to x = 5
Evaluating this integral gives:
M = (145(5) + 7.1(5^2)) - (145(0) + 7.1(0^2))
= (725 + 7.1(25)) - (0 + 0)
= 725 + 177.5
= 902.5 g
Therefore, the total mass of the rod is 902.5 grams.
b) To find the center-of-mass coordinate, we need to calculate the weighted average of the position of each infinitesimally small element of the rod, weighted by their respective mass.
The center-of-mass coordinate (x_cm) is given by the formula:
x_cm = (1/M) * ∫ (x * λ(x)) dx
Substituting the linear density function λ(x) = 145 g/m + 14.2x g/m^2 into the formula, we have:
x_cm = (1/902.5) * ∫ (x * (145 + 14.2x)) dx
Expanding and integrating this function gives:
x_cm = (1/902.5) * ∫ (145x + 14.2x^2) dx
x_cm = (1/902.5) * (72.5x^2 + 4.73x^3) ] from x = 0 to x = 5
Evaluating this integral gives us:
x_cm = (1/902.5) * [(72.5(5)^2 + 4.73(5)^3) - (72.5(0)^2 + 4.73(0)^3)]
= (1/902.5) * [(72.5(25) + 4.73(125)) - (0 + 0)]
= (1/902.5) * (1812.5 + 591.25)
= (1/902.5) * 2403.75
= 2.667 m
Therefore, the center-of-mass coordinate of the rod is 2.667 meters.
a) To determine the total mass of the rod, we need to integrate the linear density function λ(x) over the entire length of the rod.
The linear density function given is:
λ(x) = 145 g/m + 14.2x g/m^2
To find the mass at any position x, we can multiply the linear density λ(x) by the differential length dx, and integrate it over the length of the rod:
m = ∫ (λ(x) dx)
Since we have the expression for λ(x), we can substitute it into the integral:
m = ∫ (145 g/m + 14.2x g/m^2) dx
Integrating λ(x) with respect to x gives:
m = ∫145 dx + ∫14.2x dx
m = 145x + 7.1x^2 + C, where C is the constant of integration.
To find the total mass of the rod, we need to evaluate this expression between the limits x = 0 to x = 5 m:
m = 145(5) + 7.1(5)^2 + C
m = 725 + 177.5 + C
m = 902.5 + C
So, the total mass of the rod is 902.5 grams plus the constant C. We cannot determine the exact value of C without additional information.
b) To determine the center-of-mass coordinate, we need to find the average position of the mass distribution along the rod. This can be done using the concept of the centroid or by calculating the weighted average.
One way to calculate the center-of-mass coordinate is by using the formula:
x_cm = (1/m) ∫ (x * λ(x) dx)
Substituting the expression for λ(x), we have:
x_cm = (1/m) ∫ (x * (145 g/m + 14.2x g/m^2) dx)
Simplifying:
x_cm = (1/m) [(145/m) ∫ x dx + (14.2/m) ∫ x^2 dx]
Integrating both terms:
x_cm = (1/m) [(145/m) (x^2/2) + (14.2/m) (x^3/3)] + C2, where C2 is the constant of integration.
To find the center-of-mass coordinate, we evaluate this expression over the entire length of the rod, x = 0 to x = 5 m:
x_cm = (1/m) [(145/m) (5^2/2) + (14.2/m) (5^3/3)] + C2
x_cm = (1/m) [(145/m) (25/2) + (14.2/m) (125/3)] + C2
x_cm = (1/m) [(145/m) (12.5) + (14.2/m) (41.67)] + C2
So, the center-of-mass coordinate is equal to [(145/m) (12.5) + (14.2/m) (41.67)] plus the constant C2. We cannot determine the exact value of C2 without additional information.