A copper block rests 29.3 cm from the center of a steel turn

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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

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

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