A long thin rod lies along the x-axis from the origin to x=L, with L= 0.750 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x3). The value of λ0 is 0.600 kg/m and x is in meters.
1) Calculate the total mass of the rod.
(in kg)
2) Calculate the x-coordinate of the center of mass of the rod.
(in m)
3) Calculate the moment of inertia of the rod with respect to the y-axis.
(in kg*m^2)
1) To calculate the total mass of the rod, we need to integrate the mass per unit length λ along the x-axis from x=0 to x=L.
∫(λ dx) = ∫(λ0 (1+1.410x^3) dx) from 0 to L
Using the power rule for integration, we can integrate the expression inside the integral:
∫(λ0 (1+1.410x^3) dx) = λ0 * (x + 0.3525x^4) from 0 to L
Now, we can substitute the values of λ0 and L into the equation:
Total mass = λ0 * (L + 0.3525L^4) - λ0 * (0 + 0.3525(0)^4)
Total mass = λ0 * (L + 0) = λ0 * L
Plugging in the values, we get:
Total mass = 0.600 kg/m * 0.750 m = 0.450 kg
So, the total mass of the rod is 0.450 kg.
2) To calculate the x-coordinate of the center of mass of the rod, we can use the formula:
xcm = ∫(x * λ dx) / ∫(λ dx) from 0 to L
First, let's calculate the numerator:
∫(x * λ dx) = ∫(x * λ0 (1+1.410x^3) dx) from 0 to L
Using the power rule for integration, we can integrate the expression inside the integral:
∫(x * λ0 (1+1.410x^3) dx) = λ0 * (∫(x + 1.410x^4) dx) from 0 to L
∫(x + 1.410x^4) dx = (1/2)*x^2 + (1.410/5)*x^5 + C
Now, let's substitute the values of λ0 and L:
∫(x + 1.410x^4) dx = (1/2)*L^2 + (1.410/5)*L^5 - (1/2)*0^2 - (1.410/5)*0^5
Simplifying further:
∫(x + 1.410x^4) dx = (1/2)*L^2 + (1.410/5)*L^5
Now, let's calculate the denominator:
∫(λ dx) = ∫(λ0 (1+1.410x^3) dx) from 0 to L
Using the power rule for integration, we can integrate the expression inside the integral:
∫(λ0 (1+1.410x^3) dx) = λ0 * (x + 0.3525x^4) from 0 to L
∫(λ0 (1+1.410x^3) dx) = λ0 * (L + 0.3525L^4) - λ0 * (0 + 0.3525(0)^4)
Simplifying further:
∫(λ0 (1+1.410x^3) dx) = λ0 * L
Now, let's calculate the center of mass using these values:
xcm = (∫(x * λ dx)) / (∫(λ dx)) = (λ0 * (∫(x + 1.410x^4) dx)) / (λ0 * L)
xcm = (∫(x + 1.410x^4) dx) / L
Plugging in the values we have already calculated:
xcm = ((1/2)*L^2 + (1.410/5)*L^5) / L
Simplifying further:
xcm = (1/2)*L + (1.410/5)*L^4
Plugging in the value of L, we get:
xcm = (1/2)*0.750 + (1.410/5)*(0.750)^4
Calculating this expression gives us:
xcm ≈ 0.375 + 0.986
xcm ≈ 1.361 m
So, the x-coordinate of the center of mass of the rod is approximately 1.361 m.
3) To calculate the moment of inertia of the rod with respect to the y-axis, we can use the formula:
Iy = ∫(y^2 * λ dx) from 0 to L
But since the rod is lying along the x-axis, the distance from the y-axis to any point on the rod is always zero. Therefore, the moment of inertia with respect to the y-axis would be zero.
Iy = 0 kg*m^2
So, the moment of inertia of the rod with respect to the y-axis is 0 kg*m^2.
To find the total mass of the rod, we need to integrate the mass per unit length function λ(x) from x = 0 to x = L.
Given:
λ(x) = λ0 (1 + 1.410x^3)
λ0 = 0.600 kg/m
L = 0.750 m
1) Total mass of the rod:
To find the total mass:
M = ∫λ(x) dx from x = 0 to x = L
M = ∫(λ0 (1 + 1.410x^3)) dx from x = 0 to x = L
M = λ0 ∫(1 + 1.410x^3) dx from x = 0 to x = L
M = λ0 [x + (1.410/4)x^4] from x = 0 to x = L
M = λ0 (L + (1.410/4)L^4) - λ0 (0 + (1.410/4)0^4)
M = λ0 (L + (1.410/4)L^4)
Substituting the given values:
λ0 = 0.600 kg/m
L = 0.750 m
M = (0.600)(0.750 + (1.410/4)(0.750)^4)
M = 0.600(0.750 + (1.410/4)(0.125))
M = 0.600(0.750 + 0.001365234375)
M = 0.600(0.751365234375)
M = 0.450819140625 kg
Therefore, the total mass of the rod is approximately 0.451 kg.
2) Center of mass of the rod:
The x-coordinate of the center of mass is given by the equation:
x_cm = (1/M) ∫(x * λ(x)) dx from x = 0 to x = L
Substituting the values:
M = 0.450819140625 kg
λ(x) = λ0 (1 + 1.410x^3)
λ0 = 0.600 kg/m
L = 0.750 m
x_cm = (1/0.450819140625) ∫(x * 0.600 (1 + 1.410x^3)) dx from x = 0 to x = 0.750
x_cm = (2.215032413) ∫(x + 1.410x^4) dx from x = 0 to x = 0.750
x_cm = (2.215032413) [ (1/2)x^2 + (1.410/5)x^5 ] from x = 0 to x = 0.750
x_cm = (2.215032413) [ (1/2)(0.750)^2 + (1.410/5)(0.750)^5 ]
x_cm = (2.215032413) [ (1/2)(0.5625) + (1.410/5)(0.10546875) ]
x_cm = (2.215032413) [ 0.28125 + 0.029025 ]
x_cm = (2.215032413) [ 0.310275 ]
x_cm = 0.686852688711 m
Therefore, the x-coordinate of the center of mass of the rod is approximately 0.687 m.
3) Moment of inertia of the rod with respect to the y-axis:
The moment of inertia of a rod with respect to the y-axis is given by the equation:
I = ∫(λ(x) * x^2) dx from x = 0 to x = L
Substituting the values:
λ(x) = λ0 (1 + 1.410x^3)
λ0 = 0.600 kg/m
L = 0.750 m
I = ∫(0.600 (1 + 1.410x^3) * x^2) dx from x = 0 to x = 0.750
I = 0.600 ∫(x^2 + 1.410x^5) dx from x = 0 to x = 0.750
I = 0.600 [ (1/3)x^3 + (1.410/6)x^6 ] from x = 0 to x = 0.750
I = 0.600 [ (1/3)(0.750)^3 + (1.410/6)(0.750)^6 ]
I = 0.600 [ (1/3)(0.421875) + (1.410/6)(0.10546875)^2 ]
I = 0.600 [ 0.140625 + 0.010257081 ]
I = 0.600 [ 0.150882081 ]
I = 0.0905292486 kg*m^2
Therefore, the moment of inertia of the rod with respect to the y-axis is approximately 0.091 kg*m^2.
To solve these problems, we can split the rod into small segments and integrate over the entire length to find the desired quantities.
1) To calculate the total mass of the rod, we need to find the mass per unit length at each point and integrate it over the length of the rod.
The equation for the mass per unit length is given by λ = λ0 (1+1.410x^3).
We need to integrate this equation from x=0 to x=L. The limits of integration are determined by the length of the rod, which is L = 0.750 m.
The total mass of the rod, M, is obtained by integrating λ with respect to x, multiplied by the infinitesimal length dx.
M = ∫(λ dx) from 0 to L
Substituting the equation for λ, we have:
M = ∫(λ0(1+1.410x^3) dx) from 0 to L.
Integrating this expression, we get:
M = λ0∫(1+1.410x^3) dx from 0 to L.
Integrating term by term, we have:
M = λ0[x + 0.353x^4] evaluated from 0 to L.
Plugging in the limits of integration, we get:
M = λ0[L + 0.353L^4 - 0 - 0.353(0)^4].
Simplifying this expression gives:
M = λ0[L + 0.353L^4].
Now, we substitute the value of λ0, which is 0.600 kg/m, and the length of the rod, which is 0.750 m:
M = 0.600[0.750 + 0.353(0.750)^4].
Evaluating this expression gives us the total mass of the rod.
2) The x-coordinate of the center of mass of the rod, denoted as x_cm, can be calculated using the formula:
x_cm = (1/M)∫(x λ dx) from 0 to L.
To find the center of mass, we need to integrate the product of position (x) and mass per unit length (λ) with respect to x, over the length of the rod.
Using the equation for λ as given, the equation becomes:
x_cm = (1/M)∫(x λ0(1+1.410x^3) dx) from 0 to L.
Simplifying this expression, we have:
x_cm = (1/M)∫(x λ0 + 1.410x^4λ0) dx from 0 to L.
Taking out the constants, we get:
x_cm = (λ0/M)∫(x + 1.410x^4) dx from 0 to L.
Integrating term by term, we have:
x_cm = (λ0/M)[x^2/2 + 0.282x^5/5] evaluated from 0 to L.
Substituting the limits of integration, we have:
x_cm = (λ0/M)[L^2/2 + 0.282L^5/5 - 0 - 0.282(0)^5/5].
Simplifying this expression, we get:
x_cm = (λ0/M)[L^2/2 + 0.282L^5/5].
Now, we substitute the values of λ0 and M:
x_cm = (0.600/M)[L^2/2 + 0.282L^5/5].
Finally, substitute the value of M (which we calculated in part 1), and evaluate this expression to find the x-coordinate of the center of mass.
3) To calculate the moment of inertia of the rod with respect to the y-axis, denoted as I_y, we integrate the product of mass per unit length (λ) and the square of the distance from the y-axis (x^2), over the length of the rod.
I_y = ∫(x^2 λ dx) from 0 to L.
Using the equation for λ as given, the equation becomes:
I_y = ∫(x^2 λ0 (1+1.410x^3) dx) from 0 to L.
Simplifying this equation, we have:
I_y = λ0∫(x^2 (1+1.410x^3) dx) from 0 to L.
Expanding the expression inside the integral, we get:
I_y = λ0∫(x^2 + 1.410x^5) dx from 0 to L.
Taking out the constant terms, we have:
I_y = λ0∫(x^2 dx + 1.410x^5 dx) from 0 to L.
Integrating term by term, we have:
I_y = λ0[x^3/3 + 0.282x^6/6] evaluated from 0 to L.
Substituting the limits of integration, we have:
I_y = λ0[L^3/3 + 0.282L^6/6 - 0 - 0.282(0)^6/6].
Simplifying this expression, we get:
I_y = λ0[L^3/3 + 0.282L^6/6].
Finally, substitute the value of λ0 and evaluate this expression to find the moment of inertia of the rod with respect to the y-axis.