Work done and rotation of record player

[mattyisphtty]

Homework Statement

An object of mass m sits at a distance R form the axis of rotation of a record that is being played. The record makes a complete revolution in time T and friction holds the object in its place relative to the record. The coefficient of static friction between the object and the record is uS. How much work is done by friction in one revolution?

Homework Equations

W = F*d
Force friction = uS * normal force
Not sure if this is relevant but circular motion is as follows
a = v^2 / r
v = 2 pi r / T

The Attempt at a Solution

Using Frictional force eqn i got
W = uS * mg * 2piR
Using Circular motion I got
W = (8 pi^3 r^2 m)/ t^2

 

[mattyisphtty]

Would be much appreciated if someone could reply to me at least with which method to use, me and a few of my friends are working and unfortunatly we all keep getting different answers.

 

[mattyisphtty]

Futher insight: it is a multiple choice problem
A. None
B. (4 pi^2mR) / T^2
C. (8pi^3mR^2)/T^2
D. 2uSmgR
E. (4pi^2uSmgR)/T^2

although I have come across a thought that surprised me. Under these conditions you wouldn't be able to use uS unless it was on the breaking point between uS and uK? so wouldn't that rule out going the frictional force method and leaving me with the circular motion method?

 

[mattyisphtty]

Could I please get some help on this? Still pondering and yet to come up with a definite result.