一道英文物理题,主要不太懂I怎么选A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kgm2.If the string is pulled with a force,the resulting angular acceleration of the pulley is 2 rad/s2.Determine the magnitu
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![一道英文物理题,主要不太懂I怎么选A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kgm2.If the string is pulled with a force,the resulting angular acceleration of the pulley is 2 rad/s2.Determine the magnitu](/uploads/image/z/6998996-20-6.jpg?t=%E4%B8%80%E9%81%93%E8%8B%B1%E6%96%87%E7%89%A9%E7%90%86%E9%A2%98%2C%E4%B8%BB%E8%A6%81%E4%B8%8D%E5%A4%AA%E6%87%82I%E6%80%8E%E4%B9%88%E9%80%89A+string+is+wrapped+around+a+pulley+of+radius+0.05+m+and+moment+of+inertia+0.2+kgm2.If+the+string+is+pulled+with+a+force%2Cthe+resulting+angular+acceleration+of+the+pulley+is+2+rad%2Fs2.Determine+the+magnitu)
一道英文物理题,主要不太懂I怎么选A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kgm2.If the string is pulled with a force,the resulting angular acceleration of the pulley is 2 rad/s2.Determine the magnitu
一道英文物理题,主要不太懂I怎么选
A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kgm2.If the string is pulled with a force,the resulting angular acceleration of the pulley is 2 rad/s2.Determine the magnitude of the force.
A.\x050.4 N\x05\x05 \x05
B.\x052 N\x05\x05 \x05
C.\x058 N\x05\x05 \x05
D.\x0516 N\x05 \x05
E.\x0540 N
只有这么一个表,如何决定转动惯量到底是1/2MR^2还是2/3MR^2还是MR^2
他给的太乱我不太会区分啊
一道英文物理题,主要不太懂I怎么选A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kgm2.If the string is pulled with a force,the resulting angular acceleration of the pulley is 2 rad/s2.Determine the magnitu
题目翻译如下:A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kgm2.翻译过来大概是一根绳子缠绕在一个半径为0.05惯量为0.2的滑轮上,
If the string is pulled with a force,the resulting angular acceleration of the pulley is 2 rad/s2.Determine the magnitude of the force.如果绳子用力拉会导致滑轮产生一个为2的角加速度,求力的大小
图片上的公式依次是:细棒,圆盘和实体圆柱,圆环和空心圆柱,实心球,空心球的惯量公式
根据题目来选
滑轮应该根据圆盘来计算吧?!(这一点我说实话不确定,因为滑轮也有实心的类似空心的,我认为这里会偏实心),所以公式选择圆盘和圆柱体的惯量公式就是1/2MR^2,
此处会用到公式:角加速度×半径 = 线加速度即A=aR.
1/2MR^2=0.2,线加速度A=aR,F=MA=aRM
其中M=0.2*2/R,带入可得F=16N
ps:刚在百度搜了搜,滑轮是安以上所说的公式处理,如下:滑轮在滑动过程中会产生转动惯量Mr^2/2