Answer:
a. 16 slug b. 3.2 ft
Explanation:
a. Total mass of the rod
Since the linear density at a point of the rod,λ varies directly as the third power of the measure of the distance of the point form the end, x
So, λ ∝ x³
λ = kx³
Since the linear density λ = 2 slug/ft at then center when x = L/2 where L is the length of the rod,
k = λ/x³ = λ/(L/2)³ = 8λ/L³
substituting the values of the variables into the equation, we have
k = 8λ/L³
k = 8 × 2/4³
k = 16/64
k = 1/4
So, λ = kx³ = x³/4
The mass of a small length element of the rod dx is dm = λdx
So, to find the total mass of the rod M = ∫dm = ∫λdx we integrate from x = 0 to x = L = 4 ft
M = ∫₀⁴dm
= ∫₀⁴λdx
= ∫₀⁴(x³/4)dx
= (1/4)∫₀⁴x³dx
= (1/4)[x⁴/4]₀⁴
= (1/16)[4⁴ - 0⁴]
= (256 - 0)/16
= 256/16
= 16 slug
b. The center of mass of the rod
Let x be the distance of the small mass element dm = λdx from the end of the rod. The moment of this mass element about the end of the rod is xdm = λxdx = (x³/4)xdx = (x⁴/4)dx.
We integrate this through the length of the rod. That is from x = 0 to x = L = 4 ft
The center of mass of the rod x' = ∫₀⁴(x⁴/4)dx/M where M = mass of rod
= (1/4)∫₀⁴x⁴dx/M
= (1/4)[x⁵/5]₀⁴/M
= (1/20)[x⁵]₀⁴/M
= (1/20)[4⁵ - 0⁵]/M
= (1/20)[1024 - 0]/M
= (1/20)[1024]/M
Since M = 16, we have
x' = (1/20)[1024]/16
x' = 64/20
x' = 3.2 ft