How Do Velocity and Friction Affect a Coin's Motion on a Table?

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Homework Help Overview

The problem involves a coin projected forward on a table, requiring analysis of its translational and rotational motion under the influence of friction. The objective is to determine the necessary initial velocities for the coin to come to rest after traveling a specified distance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between translational and rotational motion, exploring the equations of motion and the effects of friction. There are attempts to clarify the use of variables, particularly regarding time and torque, and to derive expressions for both linear and angular velocities.

Discussion Status

Some participants have raised questions about variable usage and the assumptions made in the equations. There is an ongoing exploration of the implications of these assumptions, particularly regarding the relationship between distance, time, and velocity. A suggestion to replace certain variables has been made, but the discussion remains open without a definitive conclusion.

Contextual Notes

Participants are navigating potential confusion arising from variable notation and the assumptions about motion under friction. The problem setup involves specific conditions related to the coin's motion and the effects of kinetic friction, which are being critically examined.

Fascheue

Homework Statement



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.

Homework Equations



I = 1/2mr^2

T = r x f

Ff = uFn

F = ma

T = Iw’

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
 
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Fascheue said:
v - ug(t) = 0

v - ug(d/v) = 0
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.
 
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kuruman said:
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.
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
 
Fascheue said:
Can I replace t with 2d/v?
Yes, you can do that.
 
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