# Another friction problem (rotational this time)

1. Jun 20, 2009

### groundhog92

1. The problem statement, all variables and given/known data
Once again I just need someone to check my work, I think I have it correct. If a coin of mass .005kg is placed on a spinning record .14m from the center and it takes the coin 1.5 sec to complete a revolution then a) what is the linear speed of the coin. b)If the coefficient of static friction is .5 then at what speed would the coin slide off c) If you were to add another coin directly atop the first how would the affect your results in part b) and d) draw and label the linear velocity and instantaneous acceleration in a diagram.

2. Relevant equations
F=(mv^2)/r
Ff=umg u=coefficient of friction
Circumference/ Time = linear velocity

3. The attempt at a solution
part a
.88/1.5 = .59m/s

part b
(mv^2)/r=umg
(v^2)/r=ug
v=sqrt(rug)
v=.84m/s

part c
It wouldn't change because the masses canceled out

part d
This one I was a bit confused about. I drew the linear velocity tangent to the coins circular trajectory. I drew the instantaneous acceleration perpendicular to that going out, through the coin, from the center of the record.

Note: In the calculations for part b gravity was 10 instead of 9.81

2. Jun 20, 2009

### LowlyPion

The centripetal acceleration of the object is inward isn't it?