作业帮关于力矩torque的物理题
来源：学生作业帮 编辑：拍题作业网作业帮 分类：英语作业 时间：2024/06/05 05:07:43
关于力矩torque的物理题
A string is wrapped around a uniform disk of mass M = 1.5 kg and radius R = 0.11 m.(Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.19 m,each with a small mass m = 0.6 kg at the end.The device is initially at rest on a nearly frictionless surface.Then you pull the string with a constant force F = 23 N.At the instant when the center of the disk has moved a distance d = 0.049 m,a length w = 0.024 m of string has unwound off the disk.
(a) At this instant,what is the speed of the center of the apparatus?
(b) At this instant,what is the angular speed of the apparatus?
(c) You keep pulling with constant force 23 N for an additional 0.042 s.Now what is the angular speed of the apparatus?
绳子缠绕在一个标准圆盘上,圆盘质量为1.5kg,半径为0.11m（标准圆盘惯性为（1/2MR^2））.四个质量很小,半径为0.19m的小杆连接在圆盘上,每个小杆的末尾有一个小砝码,质量为0.6kg.这个装置最初静止在一个不记摩擦力的平面上.然后你用一个恒定的力,F=23N,拉动它.当圆盘中心移动0.049m,绳子展开量为0.024m时：
（a）这个仪器中心速度是多少
（b）这个仪器的角速度是多少
（c）在此后的0.042s中,你仍然保持用23N的力拉动它.现在它的角速度是多少?
A string is wrapped around a uniform disk of mass M = 1.5 kg and radius R = 0.11 m.(Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.19 m,each with a small mass m = 0.6 kg at the end.The device is initially at rest on a nearly frictionless surface.Then you pull the string with a constant force F = 23 N.At the instant when the center of the disk has moved a distance d = 0.049 m,a length w = 0.024 m of string has unwound off the disk.
(a) At this instant,what is the speed of the center of the apparatus?
(b) At this instant,what is the angular speed of the apparatus?
(c) You keep pulling with constant force 23 N for an additional 0.042 s.Now what is the angular speed of the apparatus?
绳子缠绕在一个标准圆盘上,圆盘质量为1.5kg,半径为0.11m（标准圆盘惯性为（1/2MR^2））.四个质量很小,半径为0.19m的小杆连接在圆盘上,每个小杆的末尾有一个小砝码,质量为0.6kg.这个装置最初静止在一个不记摩擦力的平面上.然后你用一个恒定的力,F=23N,拉动它.当圆盘中心移动0.049m,绳子展开量为0.024m时：
（a）这个仪器中心速度是多少
（b）这个仪器的角速度是多少
（c）在此后的0.042s中,你仍然保持用23N的力拉动它.现在它的角速度是多少?
装置的转动惯量为盘与四个小砝码的总和：
J=MR²/2+4*mb²=1.5*0.11²/2+4*0.6*0.19²=0.095715kgm²
总质量为1.5+4*0.6=3.9kg
(a)应用牛顿第二定律,F=(m+M)a a=F/(m+M)=23/3.9=5.90m/s²
v²=2as=2*5.9*0.049= 0.5782
v= 0.76m/s
(b)v=at t=v/a=0.1289s
角加速度为FR/J=23*0.11/0.095715=26.43 rad/s²
角速度ω=26.43*0.1289=3.41 rad/s
(c) 3.41+0.042*26.43=4.52 rad/s
J=MR²/2+4*mb²=1.5*0.11²/2+4*0.6*0.19²=0.095715kgm²
总质量为1.5+4*0.6=3.9kg
(a)应用牛顿第二定律,F=(m+M)a a=F/(m+M)=23/3.9=5.90m/s²
v²=2as=2*5.9*0.049= 0.5782
v= 0.76m/s
(b)v=at t=v/a=0.1289s
角加速度为FR/J=23*0.11/0.095715=26.43 rad/s²
角速度ω=26.43*0.1289=3.41 rad/s
(c) 3.41+0.042*26.43=4.52 rad/s