Question

一道AP 2012 力学题A ring of mass M,radius R,and rotational inerti

Answer 1

1) Mdv/dt = -μMg

积分加初始条件得：v = -μgt + v0

2) μMgR = MR^2 dw/dt

积分加初始条件w0 = 0 得：w = μgt/R

3) 因为运动距离达到L时,再无滑动,所以 wR = v,由此解得,

μgt = -μgt + v0

t = v0/(2μg)

4) 积分 v = -μgt + v0 两边从t = 0 到 t = v0/(2μg) 可得,

L = -(1/2)μgt^2 + v0 t = 3v0^2/(8μg)

美国高中数学物理教师