Frictional coin sliding on turntable

[Umphreak89]

A 5.0 g coin is placed 22 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of µs = 0.80 and µk = 0.50. What is the maximum rpm that the turntable could speed up to without the coin sliding off?



m = .005 kg
r = .22 m
µs = 0.8

Equations found..
v = angular velocity * r
Force(net) = m(v)^2 / r

Inertia > µs N when coin slips (?)

I believe this gets set equal to mg (Normal Force) but I haven't been able to generate the correct answer multiplying µs as a coefficient of either side.

I'm not sure what isn't being accounted for, what do I do?

 

[learningphysics]

Looks like you have the right idea. slipping occurs when force > us*N.

can you show your calculations for angular velocity?

 

[Umphreak89]

m(ω * r)^2 / r

simplifies to => m(r)(ω)^2

so,
mg * µs = m(r)(ω)^2 (?)

 

[learningphysics]

m(ω * r)^2 / r

simplifies to => m(r)(ω)^2

so,
mg * µs = m(r)(ω)^2 (?)

yes, solve for w. what do you get? you have to convert to rpm afterwards also.

 

[Umphreak89]

(.005)(9.8)(.8) = (.005)(.22)(ω)^2
=> (9.8)(.8) = (.22)(ω)^2
=> 7.84 / .22 = ω^2
=> ω = (7.84 / .22)^(1/2) = 5.969

then I'm not sure how to convert this to rpms..
but, 5.969 / r = 27.134, where r = .22, which isn't right

 

[learningphysics]

(.005)(9.8)(.8) = (.005)(.22)(ω)^2
=> (9.8)(.8) = (.22)(ω)^2
=> 7.84 / .22 = ω^2
=> ω = (7.84 / .22)^(1/2) = 5.969

then I'm not sure how to convert this to rpms..

first convert to revolutions/second. 1 revolution = 2*pi radians.

then convert to revolutions/minute by multiplying by 60.