m = integral fro 0 to L of λ dx
m= λo [ L + (1/3L^2)L^3 ]
m= λo [ 4L/3 ]
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Xcg = (1/m) λ x dx
= (1/m) (x^4/4L^2 + x^2/2)λo
= (1/m) (L^2)(3/4)λo
= (3/4L)(3/4)L^2 = (9/16)L whew
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for c you can do the integrals brute force or you can use the parallel axis theorem.
move to I about center of mass by subtracting m[9L/16]^2
then add m[7L/16]^2
A rod of length L has a varying density along its length that satisfies:
λ(x) =((x^2/L^2)+1)λo
where x = 0 is one end of the rod (which has density λo), and x = L is the other end (which has density 2λo).
a. Find mass of the rod in terms of L and λo.
b. Show that the center of mass of the rod, measured from the x = 0 end, is 9L/16.
c. One can show that the rotational inertia of this rod when pivoted around the end at x = 0 is 2ML^2/5. Find its rotational inertia when pivoted around its other end.
2 answers
what will be the final answer of part c?
1 answer