A long thin rod lies along the x-axis from the origin to x=L, with L= 0.830 m. The mass per unit length, $\lambda$ (in kg/m) varies according to the equation $ \lambda = \lambda_0 (1+1.110x^3). The value of $\lambda_0$ is 0.200 kg/m and x is in meters. Calculate the total mass of the rod.
- So I tried:
lambda = m/x
m = (lambda)(x)
= (0.200kg/m)(0.830m)
= 0.166kg
...which was apparently wrong so then I retried this time using the given equation:
$ \lambda = \lambda_0 (1+1.110x^3).
= (0.200kg/m)(1+(1.110)(0.830m)^3)
= 0.271kg
But that is also incorrect and I don't know what else to do.
this is integral calculus
the mass of a segment of the rod is
... segment length * lambda
lambda = .200 (1 + 1.110 x^3)
... = .200 + .222 x^3
segment length is dx
integrate from x = 0 to .83
.200 dx + .222 x^3 dx ... sum the two integrals
.166 + .026
I have no idea because I'm not in your grade. SORRY!
dM=dx(0.200*(1+1.110x^3))
taking integral on both sides,
M=0.200(L+1.110*x^4/4)
put L=0.830
M=0.19234
To calculate the total mass of the rod, we need to integrate the mass per unit length, λ, along the length of the rod. In this case, since the mass per unit length varies with x, we need to integrate λ as a function of x.
Given that λ = λ0 (1 + 1.110x^3), where λ0 = 0.200 kg/m, we can rewrite the expression for the total mass, M, as an integral:
M = ∫[0 to L] λ dx
Substituting the expression for λ:
M = ∫[0 to L] λ0 (1 + 1.110x^3) dx
Now, let's perform the integration step by step:
M = λ0 ∫[0 to L] (1 + 1.110x^3) dx
= λ0 [∫[0 to L] 1 dx + 1.110 ∫[0 to L] x^3 dx]
The first integral, ∫[0 to L] 1 dx, simply gives the length of the rod, L.
M = λ0 [L + 1.110 ∫[0 to L] x^3 dx]
To calculate the second integral, let's apply the power rule of integration:
∫ x^n dx = (1/(n+1)) x^(n+1) + C
In this case, n = 3:
∫ x^3 dx = (1/(3+1)) x^(3+1) + C
= (1/4) x^4 + C
Now we can substitute this result into the expression for the total mass:
M = λ0 [L + 1.110 (1/4) x^4) {evaluated from 0 to L}]
= λ0 [L + (1.110/4) L^4]
Finally, substitute the given values for λ0 and L:
M = (0.200 kg/m) [0.830 m + (1.110/4) (0.830 m)^4]
Evaluating this expression will give you the correct total mass of the rod.