# Centripetal acceleration and static friction

## Homework Statement

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?

mv^2/r = mg(mu)

## The Attempt at a Solution

I cant figure out what i'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v then solved for R.

I keep getting 6.204254889

## The Attempt at a Solution

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tiny-tim
Homework Helper
Welcome to PF!

Hi shaanyr! Welcome to PF!

(have a mu: µ and an omega: ω )
mv^2/r = mg(mu)

I cant figure out what i'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v
Nooo … that's ω, not v isn't it?

Use ω2r.

im sorry i dont understand

use

$$\omega^2/r$$ = g x $$\mu$$

Delphi51
Homework Helper
High school physics to the rescue!
Make a proportion to find the period:
33 turns/60 s = T turns/1 second
Once you have the period you can use v = circumference/period.

This is an interesting problem because r is unknown, yet needed to find v. In this kind of problem, you have to work with the letters and put in numbers only at the last stage when you have an r = formula.

Show your equations if you would like more help or a check!

tiny-tim
Homework Helper
Just got up :zzz: …
$$\omega^2/r$$ = g x $$\mu$$
No …

look at the dimensions (or the units) …

v = ωr,

so centripetal acceleration = v2/r = ω2r.