Respuesta :
Answer:
[tex]x_{cm}[/tex] = ¾ L (ρ₀ + ρ₀) / (2ρ₀ + ρ₁)
Explanation:
The center of mass of a body is defined as
Rcm =[tex]\frac{1}{M}[/tex] ∫ r dm
Where the blacks indicate vectors, Rcm is the position of the center of mass, M the total mass of the body, ‘r’ the position with respect to an origin
The density of a body is the relationship between two of its magnitudes that change constantly or vary in a known way, for this case we have
ρ = m / x
That can also be written in the form
ρ = dm / dx
dm = ρ dx
The expression they give us is
ρ₀ = ρ₀ + (ρ₁ -ρ₀) (x/L)²
dm =( ρ₀ + (ρ₁ -ρ₀) (x/L)² )dx
Replace in the center of mass equation and integrate, from the initial point x = 0 to the upper limit X = L . Since the whole system is on the x-axis, change the variable r by x (r --- x)
[tex]x_{cm}[/tex] = M⁻¹ ∫ x [ρ₀ + ( ρ₁-ρ₀) (x/L)²] dx
[tex]x_{cm}[/tex] = M⁻¹ [∫ ρ₀ x dx + I (ρ₁-ρ₀) / L² x³ dx]
[tex]x_{cm}[/tex] = M⁻¹ [ρ₀ x²/ + (ρ₁ -ρ₀) / L² x⁴/4]
[tex]x_{cm}[/tex] = M⁻¹ [ρ₀/2 (L2-0) + (ρ₀₁ -ρ₀) /4L² (L⁴-0)
[tex]x_{cm}[/tex] = M⁻¹ [ρ₀ L²/2 + (ρ₁ -ρ₀)/4 L²]
[tex]x_{cm}[/tex] = M⁻¹ L² [ρ₀/4 + ρ₁/4]
[tex]x_{cm}[/tex]= 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)/L² x²] dx
We integrate and evaluate between the limits of integration x = 0 and x = L
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 = L/3 (2ρ₀ + ρ₁)
Let's replace and simplify in the center of mass equation we have found
Xcm = L²/4 [ρ₀ + ρ₁] / [L/3 (2ρ₀ + ρ₀)/L]
Xcm = ¾ L (ρ₀ + ρ₀) / (2ρ₀ + ρ₁)
The center of mass of the rod with length L = [tex]\frac{3L}{4} [ ( po + po ) / ( 2po + p1 ) ][/tex]
Determine the center of mass of the rod
The density of the rod varies as
p(x) = po + ( p1 - po ) [tex](\frac{x}{L} )^2[/tex] . Also
Considering an Infinitesimal element with length dx and mass dm at distance x
dm = p(x) Sdx
where : S = area of rod. Hence total mass M = ∫dm
∫dm = [tex]\int\limits^L_o {p(x)} \, sdx[/tex]
Therefore :
[tex]\int\limits^L_o {p(x)} \, sdx = \int\limits^L_opo + [(p1-po)(\frac{x}{L} )^{2} ] \, sdx[/tex]
Resolving the equation above
[tex]\int\limits^L_o {p(x)} \, sdx =[/tex] [tex]SL [ \frac{2po + p1}{3} ][/tex]
Next step : calculate the center of mass
center of mass ( rcm ) = [tex]\frac{\int\limits^L_o {xdm} \, }{\int\limits^L_o {dm} \, }[/tex]
= [tex]\frac{\int\limits^L_o {xP(x)} \, sdx }{\int\limits^L_o {p(x)} \, sdx }[/tex]
To find the center of mass solve [tex]\int\limits^L_o {xp(x)} \, sdx[/tex]
[tex]\int\limits^L_o {xp(x)} \, sdx = S ( po \frac{x^2}{2} + ( p1 (\frac{x^{4} }{4L^2} )- po(\frac{x^4}{4L^2} ) ) )[/tex]
= [tex]S ( po \frac{L^2}{2} + ( p1 (\frac{L^{4} }{4L^2} )- po(\frac{L^4}{4L^2} ) ) )[/tex]
= [tex]S ( po (\frac{L^2}{4} ) + p1 \frac{L^{2} }{4})[/tex]
Therefore rcm ( center of mass of the rod )
rcm = [tex]L \frac{po(1/4)+ p1 (1/4)}{\frac{2po + p1}{3} }[/tex] = [tex]\frac{3L}{4} [ ( po + po ) / ( 2po + p1 ) ][/tex]
Hence we can conclude that the center of mass of a rod of length L is
[tex]\frac{3L}{4} [ ( po + po ) / ( 2po + p1 ) ][/tex]
learn more about center of mass : https://brainly.com/question/874205
