A copper block rests 29.3 cm from the center of a steel turntable.The coefficient ofstatic friction between block and surface is0.476.The turntable starts from rest and rotates with a constant angular acceleration of 0.49 rad/sThe acceleration of gravity is 9.8 m/s.After what interval will the block start toslip on the turntable?[ Hint:Ignore the tangential component of the acceleration.]Answer in units of s
问题描述：
A copper block rests 29.3 cm from the center of a steel turntable.The coefficient of
static friction between block and surface is
0.476.The turntable starts from rest and rotates with a constant angular acceleration of 0.49 rad/s
The acceleration of gravity is 9.8 m/s
.After what interval will the block start to
slip on the turntable?[ Hint:Ignore the tangential component of the acceleration.]
Answer in units of s
答
一块铜在距离一个钢转盘中心29.3厘米的位置. 铜块和转盘表面的静摩擦系数是0.476. 转盘从开始的静止不动,逐渐开始以0.49弧度/秒的恒定角加速度旋转.
重力加速度是9.8m/s
多少间隔(时间段?)之后铜块会开始在转盘上打滑(跟着旋转)?
提示:将切向分量的加速度忽略掉.
答案的单位是"s"(秒)
答
开始滑动的临界条件是：μmg=mw^2r,w=sqrt(μg/r)=sqrt(0.476*9.8/0.293)=4rad/s
即当角速度w大于4rad/s时开始滑动
则开始滑动的时间t=4/0.49=8.14s