A rod of length 30.0 cm has linear density (mass-per-length) given by
lambda = 50.0 g/m + 20.0 y g/m2
,
where y is the distance from one end, measured in meters. (a) What is the mass of the rod? (b) How far
from the y = 0 end is its center of mass?
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To find the mass of the rod, we need to integrate the linear density along the length of the rod.
The linear density is given by the equation:
λ = 50.0 g/m + 20.0y g/m²
We need to integrate this equation over the length of the rod to get the mass. The length of the rod is given as 30.0 cm, which is equal to 0.30 m.
(a) To find the mass, we integrate the linear density from 0 to 0.30:
m = ∫(0 to 0.30) λ dy
m = ∫(0 to 0.30) (50.0 g/m + 20.0y g/m²) dy
m = ∫(0 to 0.30) (50.0 + 20.0y) dy
m = 50.0y + 10.0y² | (0 to 0.30)
m = (50.0 * 0.30 + 10.0 * 0.30²) - (50.0 * 0 + 10.0 * 0²)
m = 15.0 + 0.90
m = 15.90 grams
Therefore, the mass of the rod is 15.90 grams.
(b) To find the center of mass, we need to find the position where the total mass is evenly distributed. This can be calculated using the equation for center of mass:
x_cm = (1/m) ∫(0 to 0.30) y * λ dy
Substituting the given expression for λ:
x_cm = (1/15.90) ∫(0 to 0.30) y * (50.0 g/m + 20.0y g/m²) dy
x_cm = (1/15.90) ∫(0 to 0.30) (50.0y + 20.0y²) dy
x_cm = (1/15.90) [(25.0y² + (20.0/3)y³)] | (0 to 0.30)
x_cm = (1/15.90) [(25.0 * 0.30² + (20.0/3) * 0.30³)] - (1/15.90) [(25.0 * 0² + (20.0/3) * 0³)]
x_cm = 0.0579 m
Therefore, the center of mass of the rod is located 0.0579 meters from the y = 0 end.