Force in circular motion: flat and banked roads

What keeps a body on a circle?

In the last chapter you met the centripetal acceleration, $\dfrac{v^2}{r}$ toward the centre. By the second law, something must supply a net force toward the centre:

This is the centripetal force. It is not a new kind of force. It is a job, and a known force does it: the tension of a string, friction on a road, a normal reaction, or gravity. Take that force away and the body flies off along the tangent, in a straight line.

Watch out. On a sharp bend in a car you feel thrown outward. No force pushes you out. Your body tries to go straight on (inertia), and the car door pushes you inward to turn you.

A car on a flat bend

On a flat road only friction from the road pushes the car toward the centre. It is static friction, since the tyres grip without slipping. It can be at most $\mu_s mg$, so

The mass cancels. A tighter bend or a wet road lowers the safe speed.

A banked bend

Tilt the road toward the centre by an angle $\theta$, and the normal reaction tilts too. Its sideways part now pushes the car toward the centre.

At the design speed no friction is needed at all:

• vertically, $N\cos\theta = mg$;
• toward the centre, $N\sin\theta = \dfrac{mv^2}{r}$.

Divide one by the other:

Go faster and the car tends to slide up the bank, so friction acts down it. Go slower and friction acts up the bank. With friction, there is a whole band of safe speeds. Explore it.

You should find that with $\mu = 0$ only the design speed is safe, and a rougher road widens the band on both sides.

A worked example

A $900\ \text{kg}$ car takes a flat bend of radius $65\ \text{m}$ at $15\ \text{m/s}$, with $\mu_s = 0.55$.

1. Force needed. $\dfrac{mv^2}{r} = \dfrac{900\ \text{kg} \times (15\ \text{m/s})^2}{65\ \text{m}} = 3120\ \text{N}$.
2. Friction available. $\mu_s mg = 0.55 \times 900\ \text{kg} \times 9.8\ \text{m/s}^2 = 4850\ \text{N}$. Enough, so the car is safe.
3. Fastest safe speed. $v_{\max} = \sqrt{0.55 \times 65\ \text{m} \times 9.8\ \text{m/s}^2} = 18.7\ \text{m/s}$, about $67\ \text{km/h}$.
4. Banking for $15\ \text{m/s}$. $\tan\theta = \dfrac{(15\ \text{m/s})^2}{65\ \text{m} \times 9.8\ \text{m/s}^2} = 0.353$, so $\theta = 19.5^\circ$.

Think first. Always ask which real force points to the centre. Never add a separate "centripetal force" to a free-body diagram.

A vertical circle

Whirl a bucket of water in a vertical circle of radius $r$. At the top, both the weight and the tension point down, toward the centre: $T + mg = \dfrac{mv^2}{r}$. The string stays taut only while $T \ge 0$, so the slowest speed at the top is $v = \sqrt{rg}$. At the bottom, $T - mg = \dfrac{mv^2}{r}$, so the tension is largest there.

Practise with new numbers

Try it in Python

The program finds the safe speed on a flat bend, and the bank angle needed for several design speeds.

Remember these

• Circular motion needs a net force toward the centre, $\dfrac{mv^2}{r}$. It is supplied by a real force.
• Flat bend: $v_{\max} = \sqrt{\mu_s rg}$.
• Banked bend, no friction: $\tan\theta = \dfrac{v^2}{rg}$.
• Vertical circle: the slowest speed at the top is $\sqrt{rg}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A $0.30\ \text{kg}$ puck on a smooth table, tied to a pin by a $0.90\ \text{m}$ string, moves in a circle at $4.0\ \text{m/s}$. What is the tension?

2. What is the fastest safe speed on a flat bend of radius $80\ \text{m}$ with $\mu_s = 0.35$? ($g = 9.8\ \text{m/s}^2$.)

3. Why are the bends of a racing track banked steeply?

4. A stone on a string is whirled in a horizontal circle and the string snaps. Which way does the stone go?