Let the length of a rod be 10 meters and the linear density of the rod ρ(x) be written in the

form ρ(x) = ax + b with x = 0 representing the left end of the rod and x = 10 representing
the right end of the rod. If the density of the rod is 2kg/m at the left end and 17 kg/m at the
right end, find the mass and the center of mass of the rod.

ρ(x) = ax + b

ρ(0) = a(0) + b = 2
Thus b = 2 and we have:
ρ(x) = ax + 2
Now,
ρ(10) = a(10) + 2 = 17
Thus a = 1.5 and we have:
ρ(x) = 1.5x + 2
Mass:
M = ∫ (1.5x + 2)dx from 0 to 10
= [0.75x^2 + 2x] from 0 to 10
= [0.75(10)^2 + 2(10) - 0]
= ____ kg

Centre of mass = (1/M) ∫ (x(1.5x + 2)dx from 0 to 10
= (1/____) ∫ (1.5x^2 + 2x)dx from 0 to 10
....you work out the rest.

To find the mass and center of mass of the rod, we need to integrate the linear density function over the length of the rod and use the formulas for mass and center of mass.

Let's start by finding the mass of the rod. The mass can be calculated by integrating the linear density function ρ(x) along the length of the rod.

The linear density function ρ(x) is given as ρ(x) = ax + b, where x represents the position along the length of the rod.

Given that ρ(x) = 2 kg/m at the left end (x = 0) and ρ(x) = 17 kg/m at the right end (x = 10), we can substitute these values into the linear density function to find the values of a and b.

At x = 0, we have ρ(0) = 2 kg/m, so we substitute x = 0 into ρ(x):
ρ(0) = a(0) + b = 2 kg/m
b = 2 kg/m

At x = 10, we have ρ(10) = 17 kg/m, so we substitute x = 10 into ρ(x):
ρ(10) = a(10) + b = 17 kg/m
10a + 2 = 17 kg/m
10a = 15 kg/m
a = 1.5 kg/m

Now we have found the values of a and b:
a = 1.5 kg/m
b = 2 kg/m

To find the mass of the rod, we integrate the linear density function ρ(x) over the length of the rod (from x = 0 to x = 10):

m = ∫[0 to 10] ρ(x) dx

Substituting ρ(x) = ax + b, we get:

m = ∫[0 to 10] (ax + b) dx
= ∫[0 to 10] (1.5x + 2) dx

Evaluating this integral, we get:

m = 1.5*(x^2/2) + 2x | [0, 10]
= 1.5*(10^2/2) + 2*10 - [1.5*(0^2/2) + 2*0]
= 75 + 20 - 0
= 95 kg

So, the mass of the rod is 95 kg.

Next, let's find the center of mass of the rod. The center of mass can be calculated using the formula:

x̄ = (1/m) * ∫[0 to 10] x * ρ(x) dx

Substituting ρ(x) = ax + b, we get:

x̄ = (1/m) * ∫[0 to 10] x * (ax + b) dx
= (1/95) * ∫[0 to 10] (ax^2 + bx) dx

Expanding the integrand, we get:

x̄ = (1/95) * ∫[0 to 10] (1.5ax^2 + 2bx) dx

Evaluating this integral, we get:

x̄ = (1/95) * [(1.5a/3)*x^3 + bx^2 | [0, 10]
= (1/95) * [(1.5a/3)*(10^3) + b*(10^2) - [(1.5a/3)*(0^3) + b*(0^2)]
= (1/95) * [1500a/3 + 100b]
= (1/95) * [(1500*(1.5)/3) + (100*2)]

Substituting the values of a and b, we get:

x̄ = (1/95) * [750/3 + 200]
= (1/95) * [250 + 200]
= (1/95) * 450
= 4.7368 meters (approximately)

Therefore, the center of mass of the rod is approximately 4.7368 meters.