rocophysics said:
your equations are wrong unless i screwed up
i have
\mu_{k}=\frac{1}{g}(\frac{v^{2}}{R}-\frac{F}{m})
EDIT: well i confirmed my equation by simplifying the units and all the units cancel out so seems like it's good, so i hope that helps.
How did you derive such relation? Would you mind explaining your steps?
@ camherokid:
I have a different line of thought:
Let initial angular speed be w = 40rad/s, radius of the wheel be r = 0.3m, mass of the wheel be M = 1kg, push-force be P = 10N, time be t = 2sec.
Speed of the centre of the wheel, v = w*r = 12m/s. {Assuming, no slipping.}
It has to stop in t = 2sec. Thus, final speed is zero.
Assuming uniform deceleration, a = (0 - v)/t = -v/t = -6m/s^2.
This deceleration will require an average force of magnitude, F = M*a = -M*v/t = -6N .
Now assuming this force comes only by friction, µ*P = -F. (As friction is in opposite direction; considering P to be positive.)
Thus, µ = M*v/(P.t) = (M*w*r)/(P*t) = 0.6
Therefore coefficient of friction, µ = 0.6