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WA M h 254& 15.5 c D www.webassign.ne t/web/Student/Assignment-Responses/submit?dep 12446945 4. 0/2.5 points l Previous Answers scalcET7 15.5.018 My Notes Find the moments of inertia Tx, TV, To for a lamina that occupies the part of the disk x2 y s 36 in the first quadrant if the 2 density at any point is proportional to the square of its distance from the origin. (Assume that the coefficient of proportionality is k.) Need Help? Read it Chat About It Submit Answer Save Progress Practice Another Version Submit Assignment Save Assignment Progress 6:26 PM 100%. 10/25/2015

Explanation / Answer

Use polar coordinates (r,) with x=rcos(), y=rsin() and dA=rddr

Ix = y²dA = [r=0,6] [=0,/2] r²sin²()(kr²) rddr

Ix = k [r=0,6] rdr [=0,/2] sin²()d

The two integrals can be evaluated independently.

[=0,/2] sin²()d = ½ [=0,/2] {1cos(2) }d

= ½ [=0,/2] { sin()cos() } = ½ { /2 } = /4

[r=0,6] rdr = (1/6)( 6) = 6

Ix = 1944k

Because of symmetry Iy = Ix

The polar moment I = r²dA = Ix + Iy = 3888k

This is easily confirmed

r²dA = [r=0,6] [=0,/2] r².kr². rddr = [r=0,6] krdr [=0,/2] d

= (k6/6)(/2) = k6/2 = as above

Ix = 1944k

   Iy = 1944k

   I= 3888k

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