帮解一道高一物理关于匀速圆周运动的问题

2024-11-24 12:27:22
推荐回答(3个)
回答1:

先破题;摩擦力大小为5N而没有给出方向,其方向可能是指向园心;也可能是背离园心的。
小物体静止时,弹簧拉力为F=K(L-L0)=20N .摩擦力也是20N(向外)
小物体作圆周运动时,若摩擦力方向还是向外(背离园心的);则
F-f=mω^2L 所以 ω=√(15/5*0.1)=√30 =5.477rad/s
若摩擦力的方向向里(指向园心);则;F+f=mω^2L;
所以 ω=√(25/5*0.1)=√50 =7.07rad/s

回答2:

弹簧拉力为20N
故惯性离心力为15或25N
F=mw2r
m=5 r=0.1
所以角速度w=根号下30或根号下50 rad/s

回答3:

弹簧弹力0.02*10^3=20N
当摩擦力方向与弹簧弹力同向时,w=sqr(Fr/m)=sqr(0.5)=0.707rad/s
当摩擦力方向与弹簧弹力反向时,w=sqr(0.3)=0.547rad/s

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