一道英文物理题!A potter's wheel (a solid,uniform disk) of mass 6.0kg and radius R spins about its central axis with 5.0 rad/s.A 2.0kg lump of clay is dropped onto the wheel and sticks at the edge,decreasing its angular speed.Find the new angul
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一道英文物理题!A potter's wheel (a solid,uniform disk) of mass 6.0kg and radius R spins about its central axis with 5.0 rad/s.A 2.0kg lump of clay is dropped onto the wheel and sticks at the edge,decreasing its angular speed.Find the new angul
一道英文物理题!
A potter's wheel (a solid,uniform disk) of mass 6.0kg and radius R spins about its central axis with 5.0 rad/s.A 2.0kg lump of clay is dropped onto the wheel and sticks at the edge,decreasing its angular speed.Find the new angular speed of the system.
A.\x053.0 rad/s\x05\x05 \x05
B.\x056.0 rad/s\x05\x05 \x05
C.\x052.0 rad/s\x05\x05 \x05
D.\x053.5 rad/s\x05\x05 \x05
E.\x052.5 rad/s
一道英文物理题!A potter's wheel (a solid,uniform disk) of mass 6.0kg and radius R spins about its central axis with 5.0 rad/s.A 2.0kg lump of clay is dropped onto the wheel and sticks at the edge,decreasing its angular speed.Find the new angul
这题的意思是一个质量为6kg 半径为R的圆盘,以角速度5 rad/s 绕中心轴旋转（圆盘转动惯量为mR^2/2 ） 将一个2kg的小土块静止粘在圆盘边缘,求此后圆盘转动的角速度
这道题用角动量守恒来做
粘的瞬间,可看做角动量守恒（以转动轴为轴）
初始角动量只有盘的转动（ mR^2/2 ）* 5（ 初始角速度 ）=末态角动量*末态角速度
末态角动量等于盘的角动量+土块的角动量=3R^2+2R^2=5R^2
所以带入得到末态角速度= 3.0 rad/s\x05
选A
这是一道关于刚体的习题！主要运用到转动惯量，角动量守恒定理！
解题如下：由公式知道：J=1/2mR^2;
所以知J=1/2*6*R^2=3R^2;
两物体碰撞后角动量守恒
即： J w=J' w'
而 J=3R^...
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这是一道关于刚体的习题！主要运用到转动惯量，角动量守恒定理！
解题如下：由公式知道：J=1/2mR^2;
所以知J=1/2*6*R^2=3R^2;
两物体碰撞后角动量守恒
即： J w=J' w'
而 J=3R^2; J'=(J+m'R^2)=5R^2; w=5.0rad/s.
所以 w'=3.0rad/s.
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