The one ring sits on a turning rotating at 29 revolutions per ni tire. The ring
ID: 1789089 • Letter: T
Question
The one ring sits on a turning rotating at 29 revolutions per ni tire. The ring is 7 cm from the axis of rotation. A)Assume the ring does not slip on the turntable, calculate the linear acceleration of the ring. B) what is the minimum value of the coefficient of static friction in this case? C) if the turntable starts from rest and undergoes a constant angular acceleration for 0.25 s before attaining it’s angular speed, what is the minimum value of the coefficient if static friction required for the ring to not slip while the turntable is accelerating?
Explanation / Answer
w = 29 x 2pi rad / 60 s
w = 3.04 rad/s
(A) a_r = w^2 r = 3.04^2 x 0.07
a_r = 0.65 m/s^2
(B) us = a_r / g = 0.066
(C) wf = wi + alpha t
alpha = 3.04/ 0.25 = 12.15 rad/s^2
a_t = alpha r = 0.85 m/s^2
a_r = 0.65 m/s^2
a = sqrt(a_r^2 + a_t^2) = 1.07 m/s^2
us = a / g = 0.11