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](/uploads/image/z/6574926-30-6.jpg?t=%E5%9C%86%E7%9B%98%E7%BB%95%E4%B8%80%E6%9D%A1%E8%BF%87%E7%9B%B4%E5%BE%84%E7%9A%84%E5%9B%BA%E5%AE%9A%E8%BD%B4%E6%97%8B%E8%BD%AC%2C%E6%B1%82%E5%85%B6%E5%8A%A8%E8%83%BD%E5%8E%9F%E9%A2%98%E6%98%AF%E8%BF%99%E6%A0%B7%E7%9A%84%3AA+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
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?