HEY CAN YALL PLS ASNWER DIS!!!!!
2 answers:
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Answer:
a) x_{cm} = L / 2 , y_{cm}= L/2, b) x_{cm} = L / 2 , y_{cm}= L/2
Explanation:
The center of mass of a body is the point where all external forces are applied, it is defined by
= 1 / M ∑
= 1 /M ∫ x dm
= 1 / M ∑
= 1 / M ∫ y dm
where M is the total body mass
Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis
a) let's start with the reference zero at the left end of the horizontal rod
let's use the concept of linear density
λ = M / L = dm / dl
since the rod is on the x axis
dl = dx
dm = λ dx
let's calculate
x_{cm} = M ∫ x λ dx = λ / M ∫ x dx
x_{cm} = λ / M x² / 2
we evaluate between the lower integration limits x = 0 and upper x = L
x_{cm} = λ / M (L² / 2 - 0)
we introduce the value of the density that is cosntnate
x_{cm} = (M / L) L² / 2M
x_{cm} = L / 2
We repeat the calculation for verilla verilla
λ = M / L = dm / dy
y_{cm} = 1 / M ∫ y λ dy
y_{cm} = λ M y² / 2
= M/L 1/M (L² - 0)
y_{cm}= L/2
b) we repeat the calculation for the origin the reference system is top of the vertical rod
horizontal rod
x_{cm} = 1 / M ∫λ x dx = λ/M x² / 2
we evaluate between the lower limits x = 0 and the upper limit x = -L
x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M
x_{cm} = L / 2
vertical rod
y_{cm} = 1 / M ∫y dm
y_{cm} = λ / M ∫y dy
y_{cm} = λ / M y2 / 2
we evaluate between the integration limits x = 0 and higher x = -L
y_{cm} = (M / L) 1 / M ((-L)²/2 -0)
y_{cm} = L / 2