Respuesta :
Answer:
(a). The mass of the rod is 15.9 g.
(b). The center of mass is 0.153 m.
Explanation:
Given that,
Length = 30.0 cm
Linear density [tex]\labda=50.0+20.0x[/tex]
We need to calculate the mass of rod
Using formula of mass
[tex]M=\int{dm}[/tex]
[tex]M=\int{(50.0+20.0x)dx}[/tex]
[tex]M=50.0x+10x^2[/tex]
Put the value of x
[tex]M=50.0\times0.30+10\times(0.30)^2[/tex]
[tex]M=15.9\ g[/tex]
We need to calculate center of mass
The center of mass has an x coordinate is given by
[tex]x_{cm}=\dfrac{\int{xdm}}{\int{dm}}[/tex]
We need to calculate the value of [tex]\int{xdm}[/tex]
[tex]\int{xdm}=\int{(50.0x+20.0x^2)dx}[/tex]
[tex]\int{xdm}=25x^2+\dfrac{20}{3}x^3[/tex]
Put the value into the formula
[tex]\int{xdm}=25\times0.3^2+\dfrac{20}{3}\times(0.3)^3[/tex]
[tex]\int{xdm}=2.43[/tex]
Put the value into the formula of center of mass
[tex]x_{cm}=\dfrac{2.43}{15.9}[/tex]
[tex]x_{cm}=0.153\ m[/tex]
Hence, (a). The mass of the rod is 15.9 g.
(b). The center of mass is 0.153 m.
A) The mass of the rod = 15.9 grams
B) The position of the center of mass = 0.153 m
Given data :
Length of rod ( x ) = 30 cm = 0.30 m
Linear density = 50.0 + 20.0x
A) Determine the mass ( m ) of the rod by integrating the Linear density
M = ∫ dm.dx
= ∫ ( 50.0 + 20.0x ) dx
∴ M = 50x + 20x² ----- ( 1 )
Input the value of x into equation ( 1 )
M = 50 ( 0.30 ) + 20 ( 0.30 )² = 15.9 grams
B) Determine the value of the position ( x ) of the center of mass
applying the formula below
x = ∫ [tex]\frac{xdm}{dm}[/tex] ----- ( 2 )
first step : calculate the value of ∫ xdm
∫ xdm = ∫ ( 50x + 20x² ) dx
= 25x² + [tex]\frac{20}{3} x^{3}[/tex] -------- ( 3 )
where ; x = 0.30 m ( input value into equation 3 )
∴ ∫ xdm = 2.43
Back to equation ( 2 )
X = 2.43 / 15.9
= 0.153 m.
Hence we can conclude that ; The mass of the rod = 15.9 grams, The position of the center of mass = 0.153 m
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