Question

A 1500-kg car moving o a flat, horizontal road negotiates a curve as show in figure. 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 and still make 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

(a) The force of static friction directed towards the centre of the curve keeps the car moving in a circular path.
(b) Free-body diagram for the car.

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

$f_{s,max}={\mu }_{s}N.$
$f_{s,max}={\mu }_{s}N=m\frac{v^{2}max}{r}......\left(i\right)$
Apply the particle in equilibrium model to the car in the vertical direction:
$\sum F_y=0\to N-mg=0\to N=mg$
Solve equation (i) for the maximum speed and substitution for N:

$v_{max}=\sqrt{\frac{{\mu }_{s}Nr}{m}}=\sqrt{\frac{{\mu }_{s}mgr}{m}}=\sqrt{{\mu }_{s}gr}.....\left(ii\right)$
=$\sqrt{\left(0.50\right)\left(10m/s^2\right)\left(20m\right)}=10.0m/s$