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A long thin rod lies along the x-axis from the origin to x=L, with L= 0.890 m. The mass per unit length, λ (in kg/m) varies according to the equation λ = λ0 (1+1.410x2). The value of λ0 is 0.700 kg/m and x is in meters.

1. Calculate the total mass of the rod.

*Here I got a correct answer: 8.549×10-1 kg

2. Calculate the x-coordinate of the center of mass of the rod.
??

3. Calculate the moment of inertia of the rod with respect to the y-axis.
???

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1 answer

  1. dm = .7 (1+1.41 x^2) dx

    m = .7 [ x + 1.41 x^3/3] 0 to .89

    = .854935 etc yes check

    cg = (integral x dm)/m

    = .7[ x + 1.41 x^3] dx /m

    = .7[x^2/2 + 1.41 x^4/4 ]/m 0 to .89

    = .43205147/.854935

    =.50536

    I = integral x^2 dm
    = integral x^2[ .7 (1+1.41 x^2) dx]
    = .7 int [x^2 + 1.41 x^4 ] dx same limits

    = .7 l [x^3/3 +1.41 x^5/5] at x = .89

    = .274722

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