Asked by phill
A coin rests 15 cm away from the center of a turntable. The turntable starts to rotate at an angular acceleration of .8 rad/s^2 and after 7 seconds it slides off. What is the coefficient of static friction between the coin and the surface?
Just give me step by step on how to solve this problem.
Just give me step by step on how to solve this problem.
Answers
Answered by
drwls
It starts sliding when the static friction force reaches its maximum value. Its maximum value is Us*M*g.
Us is the static coefficeint you want to determine
After 7 seconds, the angular velocity is 5.6 rad/s and the centripetal acceleration is
Rw^2 = -.15*(5.6)^2 = 4.7 m/s^2.
There is also continuing tangential acceleration of R*alpha = 0.12 m/s^2 The vector-sum acceleration is
4.701 m/s^2, so you can ignore the tangential acceleration effect.
Us*M*g = M R w^2
Us = R w^2/g = 4.7/9.8 = 0.48
Us is the static coefficeint you want to determine
After 7 seconds, the angular velocity is 5.6 rad/s and the centripetal acceleration is
Rw^2 = -.15*(5.6)^2 = 4.7 m/s^2.
There is also continuing tangential acceleration of R*alpha = 0.12 m/s^2 The vector-sum acceleration is
4.701 m/s^2, so you can ignore the tangential acceleration effect.
Us*M*g = M R w^2
Us = R w^2/g = 4.7/9.8 = 0.48
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