Question

An object of mass 1.4kg is placed 12.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the object remains fixed on the turntable until a rate of 36 rpm is reached, at which point the object slides off. What is the coefficient of static friction between the object and the turntable?

Answer 1

n=36rpm=36/60=0.6 rev/s
r=0.12 m
m•a=F(fr)
m•a=m•ω²•r=m• (2•π•n)² •r
F(fr)=μ•N=μ•m•g
m• (2•πvn)² •r= μ•m•g
μ=(2•πvn)² •r/g=(2π•0.6)² •0.12/9.8=0.174.