4)The measure of the linear density at a point of a rod varies directly as the third power of the measure of the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center, find the total mass of the given rod and the center of the mass​

a. 16 slug b. 3.2 ft

Step-by-step 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

the answer is 0.3175 hope that you ^w^

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