Uniform Circular motion, angular velocity and Friction

[Patty-o]

Homework Statement

A 7.00 g coin is placed 17.0 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of \mus = 0.850 and \muk = 0.540

What is the maximum angular velocity with which the turntable can spin without the coin sliding?

Homework Equations

a=m(V^2)/r

The Attempt at a Solution

I will assume that the weight and the normal force cancel each other out, so I am only dealing with the forces on the horizontal plane.

Normal force = weight = m*g = (.007g*9.8)= .0686N

I find the force of the static friction and the kinetic friction.
F_{s}=\mu*n=(.850*.0686)=.05831
F_{k}=\mu*n=(.540*.0686)=.037044

This is where I start thinking. As long as the static friction is greater than the kinetic the coin will keep its current velocity, which is zero.
So for the coin to move the kinetic friction would have to be greater than the static friction (right?)

What I assume I will do is use F=ma to find the velocity of the turntable.

So I set up the problem as:
Kinetic Friction - Static Friction = m*(v^2/r)

But I seem to be stuck right there because I am not sure how to get from the velocity to the angular velocity, as the angular velocity is the ratio in the change of the angle divided by the change in time.

Did I do the above steps correctly, and is my thinking correct as well? If so, how do I get to the angular velocity in rad/s?
I feel like there is some info missing from the problem

 

[alphysicist]

Hi Patty-o,

Homework Statement

A 7.00 g coin is placed 17.0 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of \mus = 0.850 and \muk = 0.540

What is the maximum angular velocity with which the turntable can spin without the coin sliding?

Homework Equations

a=m(V^2)/r

The Attempt at a Solution

I will assume that the weight and the normal force cancel each other out, so I am only dealing with the forces on the horizontal plane.

Normal force = weight = m*g = (.007g*9.8)= .0686N

I find the force of the static friction and the kinetic friction.
F_{s}=\mu*n=(.850*.0686)=.05831
F_{k}=\mu*n=(.540*.0686)=.037044

This is where I start thinking. As long as the static friction is greater than the kinetic the coin will keep its current velocity, which is zero.
So for the coin to move the kinetic friction would have to be greater than the static friction (right?)

No; the formula you have for the static friction should be:

<br /> f_{s,{\rm max}}=\mu_s N<br />

or

<br /> f_s \le \mu_s N<br />

because it tells you the maximum value of the static friction. You can calculate how much frictional force is required to hold the coin in place; if the required amount is greater than the static frictional force can provide then it will slip (and once it is slipping then you would start using your kinetic frictional force formula to find the friction).

What I assume I will do is use F=ma to find the velocity of the turntable.

So I set up the problem as:
Kinetic Friction - Static Friction = m*(v^2/r)

So for this equation, you would use either kinetic friction or static friction (but not both), making your choice based on whether the coin is slipping or not.

But I seem to be stuck right there because I am not sure how to get from the velocity to the angular velocity, as the angular velocity is the ratio in the change of the angle divided by the change in time.

Here is an webpage on angular quantities and relating angular and linear quantities:

http://hyperphysics.phy-astr.gsu.edu/HBASE/rotq.html

Did I do the above steps correctly, and is my thinking correct as well? If so, how do I get to the angular velocity in rad/s?
I feel like there is some info missing from the problem