centripetal acceleration / static friction?

Question

A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.2, how far from the center of the record can the coin be placed without having it slip off?

Answer 1

Record, what's that? Oh yeah, those ancient music things.

rohit's formula is good but the arithmetic is bad.

At the limiting distance, the maximum static friction will equal the centripetal force. The maximum available static friction is given by mu*N which you can rewrite as mu*M*g, where M is the mass. We don't know the mass but fortunately it will cancell out later. The force (centripetal) that must be given to the coin to keep it on its circular path is given by M*V^2/r where r is radius. And V is, let's see, distance traveled / delta time, so (using the distance and time for one revolution)
V = (2*pi*r/rev) / (1.8 sec/rev) = 3.49*r/s

So at the limit, where the coin as hanging on by its teeth,
0.2*M*9.8m/s^2 = M*(3.49*r/s)^2/r
evaluate and solve for r.

Answer 2

Coin will move when centrifugal force is sufficient to overcome static friction. Therefore CF=masss * Velocity squared divided by radius and force to overcome static friction is coefficient of static friction times weight of the object normal to the plane, then if the mass or weight of the object is known you can solve the equation for "r" distance from center of record.

CF=mv2/r and F=coeff static friction *m

then (mv2)/(.2 * m) = r

Answer 3

centrifugal force(pseudo outward force) = limiting static friction

mr w^2 = u mg => r= ug/w^2 = (0.2 * 10)/(2pi * 33.3/60)^2= 0.57 m

Answer 4

the answer is 8cm. yet I easily do no longer recognize a thanks to stumble on it! the answer to that mission became in my e book. I easily have an same question yet my homework makes use of .2 because the fee of static friction! i extremely do not recognize what equation to apply!