圆盘绕一条过直径的固定轴旋转,求其动能原题是这样的:A uniform sphere with mass 32.5 and radius 0.380 is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere.If the kinetic energy
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圆盘绕一条过直径的固定轴旋转,求其动能原题是这样的:A uniform sphere with mass 32.5 and radius 0.380 is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere.If the kinetic energy
圆盘绕一条过直径的固定轴旋转,求其动能
原题是这样的:A uniform sphere with mass 32.5 and radius 0.380 is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere.
If the kinetic energy of the sphere is 147 ,what is the tangential velocity of a point on the rim of the sphere?
我才大一,题目里说是绕过直径的轴转动,
圆盘绕一条过直径的固定轴旋转,求其动能原题是这样的:A uniform sphere with mass 32.5 and radius 0.380 is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere.If the kinetic energy
均匀球体转动惯量J=MR^2*2/5
球体动能E=J*W^2/2 =147
其中M=32.5,R=0.380,W为待求解角速度=?这个你自己计算下.
边缘的切向速度V=W*R好算吧?