tsr_username01
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hi, im self-teaching m4, and am struggling with finding centres of mass by intergration. can anyone explain why the answer to q6ai on the attached paper is what it is please? (have also added link for markscheme)

http://filestore.aqa.org.uk/subjects...W-QP-JUN06.PDF

http://filestore.aqa.org.uk/subjects...W-MS-JUN06.PDF


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physics4ever
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(Original post by tsr_username01)
hi, im self-teaching m4, and am struggling with finding centres of mass by intergration. can anyone explain why the answer to q6ai on the attached paper is what it is please? (have also added link for markscheme)

http://filestore.aqa.org.uk/subjects...W-QP-JUN06.PDF

http://filestore.aqa.org.uk/subjects...W-MS-JUN06.PDF


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have you tried splitting the semicircle into little slices vertically?
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WishingChaff
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Try taking moments about the y-axis first. To do this you must imagine breaking up the lamina into a series of vertical rectangular strips of width  \delta x . Then, what is the area and mass of each of these strips remembering that the lamina is uniform and its mass per unit area is denoted by  \rho . From this, you should be able to write down an expression for the torque (about the y-axis) caused by each strip of width  \delta x .

You should now be able to write an approximate expression for the total torque about the y-axis caused by the lamina. Then, taking the limit as  \delta x tends to zero you should get an integral formula for the torque. Using the fact that  x^2 + y^2 = r^2 allows you to substitute any expressions depending on y for a function of x. This yields the right hand side of the equation. I will leave you to figure out why that equates to the left.

(Original post by tsr_username01)
hi, im self-teaching m4, and am struggling with finding centres of mass by intergration. can anyone explain why the answer to q6ai on the attached paper is what it is please? (have also added link for markscheme)

http://filestore.aqa.org.uk/subjects...W-QP-JUN06.PDF

http://filestore.aqa.org.uk/subjects...W-MS-JUN06.PDF


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