A rod with a length

Question

L = 0.385 m and a nonuniform linear mass density rests along the y axis with one end at the origin. If the linear mass density of the rod is given by
λ = (5.00 ✕ 10−2 kg/m) + (1.50 ✕ 10−2 kg/m2)y then
a) what is the total mass of the rod in kg?
b) distance of the center of the mass of the rod from the origin?

Answer 1

For a) you have to integrate the linear mass density equation, then solve for the length given. ans=.0204 kg
I still haven't figured out b

Answer 2

For part b, the equation for center of mass should be:
x(cm)=(1/M)(integral of x dm)
so integrate the equation you used to find the mass from 0 to 0.385 (aka find the second integral of the initial equation and plug in the length for x) and divide by the mass.

Answer 3

For b.) you need to take the 2nd integral of the equation you are given from 0 to L(which is .365 for you). You then divide that number by the number you found in part a. I got this from looking at the center of mass equation.

Answer 4

.385* sorry about that