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设单位面积质量1,
得到此均质圆弧质量为:(α/(2π))*πa^2=(1/2)αa^2
显然,质心应在扇形的对称轴上,设其与圆心的距离为X
则:((1/2)αa^2)X=∫∫(a*cosα)*da*adα=∫∫(cosα)a^2dadα
(a从0到a,α从-α/2到α/2)
((1/2)αa^2)X=∫∫(cosα)a^2dadα=∫(cosα)dα ∫a^2da =2sin(α/2)*(1/3)a^3
=(2/3)sin(α/2)a^3
X=(4a/3)sin(α/2)
得到此均质圆弧质量为:(α/(2π))*πa^2=(1/2)αa^2
显然,质心应在扇形的对称轴上,设其与圆心的距离为X
则:((1/2)αa^2)X=∫∫(a*cosα)*da*adα=∫∫(cosα)a^2dadα
(a从0到a,α从-α/2到α/2)
((1/2)αa^2)X=∫∫(cosα)a^2dadα=∫(cosα)dα ∫a^2da =2sin(α/2)*(1/3)a^3
=(2/3)sin(α/2)a^3
X=(4a/3)sin(α/2)
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