4 0/2 points | Prev ous Answers Tipl eri 5.P.085 My Notes Ask You A 190 g disk s

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4 0/2 points | Prev ous Answers Tipl eri 5.P.085 My Notes Ask You A 190 g disk sits on a horizontally rotating turntable. The turntable makes exactly 2 revolutions each second. The disk is located 18 cm from the axis of rotation of the turntable. (a) What is the frictional force acting on the disk? 21.6 XN (b) The disk will slide off the turntable if it is located at a radius larger than 24 cm from the axis of rotation. What is the coefficient of static friction? Enter a number Submit Answer Save Progress

Explanation / Answer

(a) Angular velocity of turntable = angular displacement /T
w = (2*2Pi) / 1 = 12.57 rad/s
The distance of the disk from the center (r)= 18 cm
Now the force on the disk would be = mrw2
Since there is no motion of the disk henc this centripetal force must be equal to the friction force.
hence the friction force = mrw2 = 0.190*0.18*12.572 = 5.4 N
hence the friction force = 5.4 N
(b) If the disk will slide after the r = 24 cm then at this point the friction force is maximum
Friction force (fS)= mrw2 = 0.19*0.24*12.572 = 7.2 N -----------(1)
We know that the friction force is also defined by
fS = uSR
where uS is the coefficient of friction and R is the normal reaction of surface and given by
R= mg = 0.19*9.81 = 1.8639
fS= 1.8639uS
equating it with 1
1.8639uS = 7.2
uS = 3.865

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