Question

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a car can safely travel if the radius of the track is 84.0 m and the coefficient of friction is 0.42?

Answer 1

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 $F_c$. In this case, the only force that applies for that is the static frictional force $f_s$between the tires and the track. Then, we can write that:

$f_s=F_c$

And since $f_s\leq \mu N$ and $F_c=\frac{mv^{2}}{r}$, we have:

$\mu N\geq \frac{mv^{2}}{r}$

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:

$N-mg=0\\\\\\implies N=mg$

Substituting this expression for $N$ and solving for $v$, we get:

$\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}$

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

$v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s$

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

Answer 2

Answer:

$v=18.6 m/s$

Explanation:

We need to equal the friction force and the centripetal force:

$F_{f}=F_{c}$

$\mu N=ma_{c}$

• N is the normal force (N = mg)
• μ is the coefficient of friction (0.42)
• a(c) is the centripetal force (a_{c}=v^{2}/R)

$\mu mg=m*\frac{v^{2}}{R}$

$\mu g=\frac{v^{2}}{R}$

$v=\sqrt{R\mu g}$

$v=\sqrt{84*0.42*9.81}$

$v=18.6 m/s$

I hope it helps you!