Question

A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. The cylinder is at a distance of r = 12 cm from the center of the turntable. The coefficient of static friction between the bottom of the cylinder and the surface of the turntable is 0.45. What is the maximum speed vmax that the cylinder can have without slipping off the turntable?

Answer 1

given

cylinder distance from the center = r = 12cm = 0.12m

coeffiecent of static friction = µs = 0.45

maxium speed without slipping

    vmax = vµs gr

             = v0.45 * 9.8m/s^2 * 0.12m

               = 0.72m/s

Answer 2

The rotational force (centripetal) is mv^2/r = F
The frictional force which opposes the centripetal = u*m*g

The two forces are equal.
mv^2/r = u*m*g
The m's cancel out.
u = 0.45
g = 981 cm/s^2
r = 12 cm
v = ??????

v^2 / 12 cm = 0.45 * 981 cm/s^2
v^2 = 12 * 0.45 * 981
v^2 = 5297 cm^2/s^2
v = 72.87 cm/s

Answer 3

let it be T
so (2pi/T)^2* r=.45g
so T=1.03
omega=6.06
frequency=.96

Time period=1.03