Question

高一物理问题，急！！求详解，谢谢！！

Answer 1

绳刚断时与竖直方向夹角设为a，则Fcosa=mg=5N，F表示绳最大拉力
a=60°
绳刚断时，小球距圆柱形轴线为r=Lsina
速度v0满足v0^2=Fr/m=40/(3√3)m^2/s^2
mvt^2/2-mv0^2/2=mgh,vt为末速9m/s
h=(vt^2-v0^2)/(2g)=[81-40/(3√3)]/20m
H=h+Lcosa=[5.55+2/(3√3)]m=5.93m
t=√(2h/g)=√[0.81-0.4/(3√3)]s
x=v0t
R^2=x^2+r^2=v0^2*t^2+(Lsina)^2={40/(3√3)*[0.81-0.4/(3√3)]+27/4}m^2=[32.4/(3√3)+22.75]m^2
=28.99m^2
R=5.38m

Answer 2

在图里