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](/uploads/image/z/1676481-33-1.jpg?t=%E4%B8%80%E9%81%93%E8%8B%B1%E6%96%87%E7%89%A9%E7%90%86%E9%A2%98%21A+potter%27s+wheel+%28a+solid%2Cuniform+disk%29+of+mass+6.0kg+and+radius+R+spins+about+its+central+axis+with+5.0+rad%2Fs.A+2.0kg+lump+of+clay+is+dropped+onto+the+wheel+and+sticks+at+the+edge%2Cdecreasing+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 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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