Respuesta :
The center of mass of the rod of length L whose mass density changes from one end to the other quadratically is; X_cm = ³/₄L(ρ₀ + ρ₀)/(2ρ₀ + ρ₁)
What is the center of mass?
The formula for the center of mass of a body is defined as;
X_cm = (1/M) ∫r.dm
Where;
X_cm is the position of the center of mass
M is the total mass of the body
r is the position with respect to an origin
The density of a body is the relationship between two of its magnitudes that change constantly and so we have;
ρ = m/x
Applying small changes and we have;
ρ = dm / dx
dm = ρ*dx
We are given the expression;
ρ₀ = ρ₀ + (ρ₁ -ρ₀) (x/L)²
Thus;
dm = (ρ₀ + (ρ₁ -ρ₀) (x/L)²)dx
Since the whole system is on the x-axis, we integrate from the initial point x = 0 to the upper limit X = L. This gives;
X_cm = M⁻¹ ∫ x [ρ₀ + ( ρ₁-ρ₀) (x/L)²] dx
X_cm = M⁻¹ [∫ ρ₀ x dx + I (ρ₁-ρ₀) / L² x³ dx]
X_cm = M⁻¹ [ρ₀ x²/ + (ρ₁ -ρ₀) / L² x⁴/4]
X_cm = M⁻¹ [ρ₀/2 (L2-0) + (ρ₀₁ -ρ₀) /4L² (L⁴-0)
X_cm = M⁻¹ [ρ₀ L²/2 + (ρ₁ -ρ₀)/4 L²]
X_cm = M⁻¹ L² [ρ₀/4 + ρ₁/4]
X_cm = M⁻¹ L²/4 ( ρ₀ +ρ₁)
The only parameter that we don't know explicitly is the total mass, but we can look for their relationship using the concept of density.
M = ∫dm = ∫ρ.dx
M = ∫[ρ₀ + (ρ₁ -ρ₀)/L²x²] dx
We integrate between the limits of integration x = 0 and x = L to get;
M = ρ₀ x + (ρ₁ -ρ₀)/L² *x³/3
M = ρ₀ L + (ρ₁ -ρ₀)/3L² L³
M = ρ₀ L + (ρ₁-ρ₀)/3 L
M = L(ρ₀ + ρ₁/3 -ρ₀/3)
M = L(2/3 ρ₀ + ρ₁/3)
M = ¹/₃L(2ρ₀ + ρ₁)
We will replace and simplify in the center of mass equation to get;
X_cm = ¹/₄L²[ρ₀ + ρ₁] / [L/3 (2ρ₀ + ρ₀)/L]
X_cm = ³/₄L(ρ₀ + ρ₀)/(2ρ₀ + ρ₁)
Read more about Center of Mass at; https://brainly.com/question/13650080
