On a flat road
Circular Motion - OCR A-Level Physics
- On a flat road: friction between tyres and road provides centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion.. Maximum speed: $v_{max} =$ $\sqrt{\mu g r}$$where \mu$is the coefficient of friction.
- On a banked track: the horizontal component of the normal reaction provides centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion.. At the design speed, no friction is needed.
- For a banked track at $angle \theta$with no friction: $\tan\theta$ = \(v^{2}\)/(rg).
- If the car goes faster than the design speed, friction acts down the slope (towards the centre). If slower, friction acts up the slope.
- Banking is used on motorway curves, velodromes, and aircraft turns to reduce reliance on friction.
Worked Example
A car travels around a flat circular bend of radius 50 m. The coefficient of friction is 0.80. Calculate the maximum speed.
Show Solution
1
Friction provides centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion.
$F_{friction} = F_{centripetal}.$
2
$\mu mg = mv^2/r.$
3
Cancel m
$v^2 = \mu gr.$
4
$v = \sqrt{0.80 \times 9.81 \times 50}.$
5
$v = \sqrt{392.4} = 19.8 m s^{-1}.$
6
$$v \approx 20$ $m s^{-1}$ (about 44 mph).$
Answer
$$v_{max} \approx$ 20 $m s^{-1}$$