Modeling a center of mass of a thin wire. I am asked to find the moment about the x axis for a thin wire of constant density. This thin wire lies along the curve y=sqrtx and the limits for integration are x=0 and x=2.

Cael Dickerson

Cael Dickerson

Answered question

2022-11-15

Modeling a center of mass of a thin wire
I am asked to find the moment about the x axis for a thin wire of constant density. This thin wire lies along the curve y = x and the limits for integration are x = 0 and x = 2.
I know from my textbook that the moment about the x axis is: M y = y ~ d m
Because this is a thin wire, I know that I need to subdivide the wire into small segments for integration. I have the following for relevant data for each segment:
Length: d l = x d x
mass: d m = δ d l = δ x d x
It's the part about the distance of the center of mass to the x axis that I think I'm missing. I have the following:
y ~ = x
Therefore, my final integral is:
M y = y ~ d m = 0 2 δ x x d x = δ 0 2 x d x = δ 1 2 x 2 | 0 2 = δ 2
This particular problem is an odd numbered problem and so I know that I've got it incorrect. Please help me to see where I'm going wrong.

Answer & Explanation

Aiden Villa

Aiden Villa

Beginner2022-11-16Added 10 answers

Step 1
The center of mass for a wire of constant density making a curve y = f ( x ) between x = a and x = b is
y ¯ = a b d x y 1 + y 2 a b d x 1 + y 2
In the case you describe
y ¯ = 0 2 d x x 1 + 1 4 x 0 2 d x 1 + 1 4 x
The top integral is relatively simple and is equal to
0 2 d x x + 1 4 = 2 3 ( 27 8 1 8 ) = 13 6
Step 2
The bottom integral may be evaluated by making the substitution sec θ = 1 + 1 4 x to get
0 2 d x 1 + 1 4 x = 1 2 arccos 8 / 3 π / 2 d θ csc 3 θ = 1 8 ( 12 2 + log ( 17 + 12 2 ) )
Therefore your center of mass is
y ¯ = 52 / 3 12 2 + log ( 17 + 12 2 ) 0.846
Kayley Dickson

Kayley Dickson

Beginner2022-11-17Added 3 answers

Step 1
For arc-length integration
y ¯ = y 1 + y 2 d x 1 + y 2 d x = N r D r
Step 2
We can evaluate Numerator and Denominator by parametrization easier with t, max value being t m = y m / ( 2 f )
y 2 = 4 f x ; ( f = 1 4 ) ; x = f t 2 , y = 2 f t , x = 2 f t y = 2 f ;
N r = 2 f t 1 + 4 f 2 2 f t d t = 4 f 2 1 + 4 f 2 t m 3 / 3
D r = 1 + 4 f 2 2 f t d t = f 1 + 4 f 2 t m 2
Divide, simplify y ¯ = 4 f 3 t m = 2 3 y m = 2 2 3 0.942809

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