- #1
Father_Ing
- 33
- 3
- 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}##?
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}##?