A rod of length L has a varying density along its length that satisfies:

λ(x) =((x^2/L^2)+1)λo
where x = 0 is one end of the rod (which has density λo), and x = L is the other end (which has density 2λo).
a. Find mass of the rod in terms of L and λo.
b. Show that the center of mass of the rod, measured from the x = 0 end, is 9L/16.
c. One can show that the rotational inertia of this rod when pivoted around the end at x = 0 is 2ML^2/5. Find its rotational inertia when pivoted around its other end.

m = integral fro 0 to L of λ dx

m= λo [ L + (1/3L^2)L^3 ]

m= λo [ 4L/3 ]

--------------------------------
Xcg = (1/m) λ x dx

= (1/m) (x^4/4L^2 + x^2/2)λo

= (1/m) (L^2)(3/4)λo

= (3/4L)(3/4)L^2 = (9/16)L whew
-------------------------------
for c you can do the integrals brute force or you can use the parallel axis theorem.
move to I about center of mass by subtracting m[9L/16]^2
then add m[7L/16]^2

what will be the final answer of part c?

To find the mass of the rod in terms of L and λo, we need to integrate the density function across the length of the rod.

a. Mass of the rod:
The mass of an infinitesimally small portion of the rod at a distance x from the x = 0 end can be calculated using the formula:
dm = λ(x) * dx

Substituting the given density function into the above equation, we have:
dm = ((x^2/L^2) + 1)λo * dx

To find the total mass of the rod, we integrate the above equation from x = 0 to x = L:
M = ∫ dm = ∫ ((x^2/L^2) + 1)λo * dx
= λo/L^2 * ∫ (x^2 + L^2)dx
= λo/L^2 * [ (x^3/3) + L^2x ] [limits from x=0 to x=L]
= λo/L^2 * [ (L^3/3) + L^3 ]
= λo/L^2 * (4L^3/3)
= (4/3)λoL

Therefore, the mass of the rod is (4/3)λoL.

b. Center of mass of the rod:
To find the center of mass of the rod, we need to calculate the position x_cm where the mass is balanced. The center of mass can be found using the formula:

x_cm = (1/M) * ∫ x * dm

Substituting the value of dm from part a, we have:

x_cm = (1/M) * ∫ x * ((x^2/L^2) + 1)λo * dx
= λo / M * ∫ (x^3/L^2 + x) * dx
= λo / M * [ (x^4/4L^2) + (x^2/2) ] [limits from x=0 to x=L]
= λo / M * [ (L^4/4L^2) + (L^2/2) - (0/4L^2) - (0/2) ]
= λo / M * [ (L^2/4) + (L^2/2) ]
= λo / M * (3L^2/4)
= (3/4) L^2 / M

Substituting the value of M from part a, we have:

x_cm = (3/4) L^2 / ((4/3)λoL)
= (9L/16)

Therefore, the center of mass of the rod, measured from the x = 0 end, is 9L/16.

c. Rotational inertia when pivoted around the other end:
The rotational inertia (also known as moment of inertia) of the rod pivoted around its other end can be calculated using the parallel axis theorem.

The parallel axis theorem states that the rotational inertia I' around an axis parallel to and at a distance d from an axis through the center of mass (with inertia I) can be found using the formula:

I' = I + Md^2

Given that the rotational inertia I around the end at x = 0 is 2ML^2/5, we want to find the rotational inertia I' around the other end at x = L.

Using the parallel axis theorem, we have:

I' = I + ML^2

Substituting the given value of I, we have:

I' = (2ML^2/5) + ML^2
= 2ML^2/5 + 5ML^2/5
= 7ML^2/5

Therefore, the rotational inertia of the rod when pivoted around its other end at x = L is 7ML^2/5.