Car that undergoes non-uniform circular motion

AI Thread Summary
The discussion centers on the calculations of a car undergoing non-uniform circular motion, specifically addressing the frictional forces involved. The solution manual states that the resultant friction force must be less than or equal to the product of the coefficient of friction and the gravitational force, leading to the derived maximum velocity equation. Questions arise regarding the use of the less-than-or-equal-to sign for kinetic friction, which is typically constant at kN, and the interpretation of when the car is considered to be slipping or not. It is suggested that the book may be conflating static friction with kinetic friction in its analysis of the car's motion. Clarification on these points is essential for understanding the conditions under which the car maintains traction.
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Homework Statement
A car moves with a constant tangential acceleration Wt along a horizontal surface circumscribing a circle of radius R. The coefficient of sliding friction between the wheels and the surface is k. What distance will the car ride without sliding if at the initial moment of time its velocity is zero?
Relevant Equations
f ≤ kN
In the solution manual, it says that:
the resultant of friction force is ##<= kmg##, hence $$m\sqrt{\omega_t^2 + (\frac {v^2} {R})^2} <= kmg$$
and from this equation, we will get $$v^2 <= R \sqrt{(kg)^2 -\omega_t^2}$$
which will make ##v_{max}^2= R \sqrt{(kg)^2 -\omega_t^2}##
Finally, they calculate the distance by using ##s = \frac{v_{max}^2} {2 \omega_t}##

Now, my question is:
1.As far as I'm concerned, unlike static friction, kinetic friction has no maximum value; it is always equal to ##kN##. Why does the book use <= sign?
2.From my interpretation, What the book asks is that at what distance will the the car start to ride without slipping. Then, why the car is considered as not slipping when it reaches ##v_{max}##?
 
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Looks to me like they meant static friction.
 
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