a solid cylindrical flywheel is 1.2m in diameter and has a m

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a solid cylindrical flywheel is 1.2m in diameter and has a mass of 910kg.If the axle is 150mm i...
a solid cylindrical flywheel is 1.2m in diameter and has a mass of 910kg.If the axle is 150mm in diameter and the coefficient of journal friction is 0.15,find the time required for the flywheel to coast to rest from speed of 500rev/min.

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yoyo110 共回答了16个问题 | 采纳率87.5%
不清楚coast ,rest from正确含义,故仅为个人理解,仅供参考.“一个立体圆柱飞轮,直径为1.2米,质量为 910千克.假如轮轴的直径是150毫米,摩擦系数是0.15,请计算出飞轮达到500转/分钟的速度所需要的时间.”
1年前

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答案不是112.为什么不对啊!
方程是y = 10x - x^2,y = 21;
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答案是128pi/3.
你所得的答案使用的区间是x=0到x=3,正确的区间应该是x=3到x=7(当y=21时)
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V = pi*r^2*h = 512
A = 2*pi*r*h + 2*pi*r^2
h = 512/(pi*r^2)
A = 2*pi*r*512/(pi*r^2) + 2*pi*r^2 = 1024/r + 2*pi*r^2
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第一步:联立y=2+x-x^2, x + y =2可以解出来,两曲线的交点为x=0,y=2;x=2,y=0
第二步:由交点可以确定两曲线围成了一个闭合曲线,按题目要求,为这个闭合曲线绕y轴旋转所产生的类圆柱壳的体积,那么用微积分的元素分析法,可以确定圆柱壳的微分表达式,即
dV=2*pi*x*【g(x)-f(x)】dx ,其中g(x)=2+x-x^2,f(x)=2-x
第三步:对上面的圆柱壳微分量在0到2上积分即可求得最后的体积V=(8*pi)/3.