# Angular velocity and friction

1. Dec 1, 2017

### Fascheue

1. The problem statement, all variables and given/known data

A coin stands vertically on a table. It is projected forward (in the plane of itself) with speed v and angular speed w. The coefficient of kinetic friction between the coin and the table is u. What should v and w be so that the coin comes to rest (both translationally and rotationally) a distance d from where it started.

2. Relevant equations

I = 1/2mr^2

T = r x f

Ff = uFn

F = ma

T = Iw’

3. The attempt at a solution

For v:

Ff = uFn

Ff = umg

Fnet = umg

ma = umg

a = ug

v - ug(t) = 0

v - ug(d/v) = 0

v = (ugd)/v

v^2 = ugd

v = sqrt(ugd)

For w:

T = r x f

T = umgr

T = Iw’

umgr = 1/2mr^2w’

w’ = 2ug/r

w - (2ug/r)(t) = 0

w = (2ug/r)(d/v)

w = (2ug/r)(d/(sqrt(ugd)))

w = 2sqrt(ugd)/r

Last edited: Dec 1, 2017
2. Dec 1, 2017

### kuruman

Here you use t for "time" while later you use the same symbol for "torque". That could be confusing. Setting that aside, t for time cannot be replaced by d/v because the coin does not cover equal distances in equal times.

3. Dec 1, 2017

### Fascheue

I replaced the t’s that were used to represent torque with T.

Can I replace t with 2d/v?

d = avg velocity (t)

t = d/(avg velocity)

And average velocity is (vi + vf)/2 = (v +0)/2

t = d/(.5v) = 2d/v

4. Dec 1, 2017

### kuruman

Yes, you can do that.