You are designing a part for a piece of machinery. The part consists of a piece

Question

You are designing a part for a piece of machinery. The part consists of a piece of sheet metal cut as shown below. The shape of the upper edge of the part is given by y,(x), and the shape of the lower edge of the part is given by ½(x). y,(x) = h y2(x) = h where h 7.9 m and d 1 m Y(x) dm ½(x) dx You decide to find the moment of inertia of the part about that y axis first. The mass density per area for the sheet metal is 4.8 kg/m2. In order to find the moment of inertia, first you must chop the part into small mass elements, dm 's, that you know the moments of inertia for, ds. Then you must use an integral to sum up all of the ds. For the dm shown, what is the distance r from the axis of rotation (the y axis) to dm in terms of x, d, dx, and h? Enter your answer as a formula and be sure to show every multiplication sign. 1pts Submit Answer Tries 0/15 jsMath

Explanation / Answer

R = x

dA=(y_1- y_2)dx

dm = sigma(y_1- y_2)dx

dI = sigma(y_1- y_2)x2dx

Iy =integral sigma(h(x/d)1/3- h(x/d)3)x2 dx = sigma.h(3/10)x10/3 d-1/3   - sigma.h.x6/(6d3) =4.8×7.9×[0.3 - 1/6] =4.8×7.9×13.333=505.6kg m2

m= integral sigma(h(x/d)1/3- h(x/d)3) dx = sigma.h(3/4)x4/3 d-1/3   - sigma.h.x4/(4d3)

=4.8×7.9×( 0.75 - 0.25) = 18.96 kg

xcm = integral x.dm /m = integral sigma(h(x/d)1/3- h(x/d)3)xdx divided by m = (sigma.h(3/7)x7/3 d-1/3   - sigma.h.x5/(5d3) )/m= 4.8×7.9×(3/7 -0.2)/18.96 =0.457 m

Similarly you can do remaining two parts. They are even easier

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