Answer:
The maximum speed at which the car can safety travel around the track is 18.6m/s.
Explanation:
Since the car is in circular motion, there has to be a centripetal force [tex]F_c[/tex]. In this case, the only force that applies for that is the static frictional force [tex]f_s[/tex]between the tires and the track. Then, we can write that:
[tex]f_s=F_c[/tex]
And since [tex]f_s\leq \mu N[/tex] and [tex]F_c=\frac{mv^{2}}{r}[/tex], we have:
[tex]\mu N\geq \frac{mv^{2}}{r}[/tex]
Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:
[tex]N-mg=0\\\\\\implies N=mg[/tex]
Substituting this expression for [tex]N[/tex] and solving for [tex]v[/tex], we get:
[tex]\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}[/tex]
Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:
[tex]v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s[/tex]
It means that in its maximum value, the speed of the car is equal to 18.6m/s.
Answer:
[tex]v=18.6 m/s[/tex]
Explanation:
We need to equal the friction force and the centripetal force:
[tex]F_{f}=F_{c}[/tex]
[tex]\mu N=ma_{c}[/tex]
[tex]\mu mg=m*\frac{v^{2}}{R}[/tex]
[tex]\mu g=\frac{v^{2}}{R}[/tex]
[tex]v=\sqrt{R\mu g}[/tex]
[tex]v=\sqrt{84*0.42*9.81}[/tex]
[tex]v=18.6 m/s[/tex]
I hope it helps you!
