Question

A 1500-kg car moving on a flat, horizontal road negotiates a curve. If the radius of the curve is 20.0 m and the coefficient of static friction between the tires and dry pavement is 0.50, find the maximum speed the car can have so that it still makes the turn successfully.

• A. 20 m/s
• B. 30 m/s
• C. 10 m/s
• D. 40 m/s

Answer 1

The correct option is C

10 m/s

The force that enables the car to remain in its circular path is the force of static friction. (It is static because no slipping occurs at the point of contact between road and tires. If this force of static friction were zero- for example, if the car were on an icy road- the car would continue in a straight line and slide off the road). The maximum speed the car can have around the curve is the speed at which it is on the verge of skidding outward. At this point, the friction force has its maximum value .

$f_{s,max}={\mu }_{s}N=m\frac{v_{max}^2}{r}$ ............(i)

Apply the particle in equilibrium model to the car in the verticle direction:

$\sum F_y=0\to N-mg=0\to N=mg$

Solve equation (i) for the maximum speed and substitute for N:

$v_{max}=\sqrt{\frac{{\mu }_{s}Nr}{m}}=\sqrt{\frac{{\mu }_{s}mgr}{m}}=\sqrt{{\mu }_{s}gr}$ ......... (ii)

= $\sqrt{\left(0.50\right)\left(10m/s^2\right)\left(20m\right)}=10.0m/s$