A long thin rod lies along the x-axis from the origin to x=L, with L= 0.890 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x2). The value of λ0 is 0.700 kg/m and x is in meters.
1. Calculate the total mass of the rod.
2. Calculate the x-coordinate of the center of mass of the rod.
3. Calculate the moment of inertia of the rod with respect to the y-axis.
To find the total mass of the rod, we need to integrate the mass per unit length λ with respect to x from 0 to L.
1. To calculate the total mass of the rod:
The equation for λ is given as: λ = λ0 * (1 + 1.410x^2).
To find the total mass, we integrate λ with respect to x from 0 to L:
m = ∫λ dx
Substituting the given equation for λ:
m = ∫(λ0 * (1 + 1.410x^2)) dx
m = λ0 * ∫(1 + 1.410x^2) dx
Now integrate each term separately:
∫1 dx = x
∫(1.410x^2) dx = (1.410/3)x^3
m = λ0 * (x + (1.410/3)x^3) evaluated from 0 to L
m = λ0 * (L + (1.410/3)L^3)
Substitute the given values:
m = 0.700 * (0.890 + (1.410/3) * (0.890)^3)
Solve the expression to find the total mass of the rod.
2. To calculate the x-coordinate of the center of mass of the rod:
The center of mass of a rod is given by the equation:
x_cm = ∫(x * λ) dx / ∫λ dx
Substituting the given equation for λ:
x_cm = ∫(x * λ0 * (1 + 1.410x^2)) dx / ∫(λ0 * (1 + 1.410x^2)) dx
Performing the integration:
x_cm = λ0 * ∫(x * (1 + 1.410x^2)) dx / λ0 * ∫(1 + 1.410x^2) dx
x_cm = ∫(x + 1.410x^3) dx / ∫(1 + 1.410x^2) dx
x_cm = (∫x dx + 1.410∫x^3 dx) / (∫1 dx + 1.410∫x^2 dx)
Now integrate each term separately:
∫x dx = (1/2)x^2
∫x^3 dx = (1/4)x^4
∫1 dx = x
∫x^2 dx = (1/3)x^3
Substitute the integrals back into the equation:
x_cm = (1/2)x^2 + 1.410(1/4)x^4 / x + 1.410(1/3)x^3
Now evaluate x_cm from x = 0 to L:
x_cm = [(1/2)L^2 + 1.410(1/4)L^4] / [L + 1.410(1/3)L^3]
Simplify the expression and solve for the x-coordinate of the center of mass.
3. To calculate the moment of inertia of the rod with respect to the y-axis:
The moment of inertia of a rod is given by the equation:
I = ∫(λ * x^2) dx
Substituting the given equation for λ:
I = ∫(λ0 * (1 + 1.410x^2) * x^2) dx
Performing the integration:
I = λ0 * ∫(x^2 + 1.410x^4) dx
I = λ0 * (∫x^2 dx + 1.410∫x^4 dx)
Now integrate each term separately:
∫x^2 dx = (1/3)x^3
∫x^4 dx = (1/5)x^5
Substitute the integrals back into the equation:
I = λ0 * ((1/3)x^3 + 1.410(1/5)x^5)
Evaluate I from x = 0 to L:
I = λ0 * [((1/3)L^3 + 1.410(1/5)L^5) - ((1/3)(0)^3 + 1.410(1/5)(0)^5)]
Simplify the expression and solve for the moment of inertia of the rod with respect to the y-axis.