Why Rolling a Heavy Drum Is Easier Than Dragging It

Friction lets tyres grip the road and stops a parked scooter from sliding, while every turn a vehicle takes needs a force pulling it towards the centre of the curve. Often that inward force is friction itself, which is why speed limits on curves come straight from Newton's second law.

This lesson covers static, kinetic and rolling friction, the maximum safe speed on level and banked roads, and the vertical circle and conical pendulum.

**Static friction resists the start of sliding and adjusts up to a maximum of ${\mu }_{s}N$, kinetic friction opposes sliding with a nearly constant value ${\mu }_{k}N$, and rolling friction, much smaller than both, opposes rolling; friction is proportional to the normal reaction and nearly independent of contact area.

Types of friction:

- Static friction** acts between surfaces at rest relative to each other and matches the applied force up to a limit, $f_s\le {\mu }_{s}N$
- Kinetic friction acts during sliding: $f_k={\mu }_{k}N$, with ${\mu }_k<{\mu }_s$
- Rolling friction acts when a body rolls and is far smaller than sliding friction

Laws of friction. Limiting friction is proportional to the normal reaction, does not depend on the area of contact, and depends on the nature of the surfaces; a block on a slope begins to slide when $\tan \theta ={\mu }_s$.

Worked example 1 — pushing a crate. A 20 kg crate rests on a floor with ${\mu }_s=0.50$ and ${\mu }_k=0.40$, and $g=9.8$ m s${}^{-2}$:

$$N=20\times 9.8=196\text{ N}{f}_{\max }=0.50\times 196=98\text{ N}$$

A 60 N push does not move it, so static friction is exactly 60 N. A 120 N push moves it, and then

$$a=\frac{120-0.40\times 196}{20}=\frac{120-78.4}{20}=2.08{\text{ m s}}^{-2}$$

An everyday example. Workers at a godown roll heavy oil drums instead of dragging them, because rolling friction is far smaller than sliding friction.

The substance. **Static friction is not always equal to ${\mu }_{s}N$** — that is only its maximum; below the limit it takes whatever value stops sliding.

**On a level road friction alone provides the centripetal force, so $v_{\max }=\sqrt{{\mu }_{s}rg}$; on a road banked at angle $\theta$, the normal reaction helps too, and even without friction a banked road allows a safe speed of $\sqrt{rg\tan \theta }$.

Level road.** Vertically $N=mg$, and horizontally friction supplies $\frac{m{v}^2}{r}$. Since friction cannot exceed ${\mu }_{s}mg$:

$$\frac{m{v}^2}{r}\le {\mu }_{s}mg\Rightarrow v_{\max }=\sqrt{{\mu }_{s}rg}$$

Banked road without friction. The normal reaction tilts inwards: $N\cos \theta =mg$ and $N\sin \theta =\frac{m{v}^2}{r}$. Dividing:

$$\tan \theta =\frac{v^2}{rg}\Rightarrow v=\sqrt{rg\tan \theta }$$

Banked road with friction. At the maximum speed friction acts down the slope, so $N\cos \theta =mg+{\mu }_{s}N\sin \theta$ and $N\sin \theta +{\mu }_{s}N\cos \theta =\frac{m{v}^2}{r}$, which give

$$v_{\max }=\sqrt{\frac{rg({\mu }_s+\tan \theta )}{1-{\mu }_s\tan \theta }}$$

**Worked example — a curve of radius 50 m with ${\mu }_s=0.40$:**

- Level: $v_{\max }=\sqrt{0.40\times 50\times 9.8}=14.0$ m s${}^{-1}$, about 50 km h${}^{-1}$
- Banked at 15° without friction: $v=\sqrt{50\times 9.8\times 0.268}=11.5$ m s${}^{-1}$
- Banked at 15° with friction: $v_{\max }=\sqrt{\frac{50\times 9.8\times (0.40+0.268)}{1-0.40\times 0.268}}=19.1$ m s${}^{-1}$, about 69 km h${}^{-1}$

An everyday example. Tight bends on hill roads carry low speed-limit signs, because a small radius sharply reduces the maximum safe speed.

The substance. The safe speed does not depend on the vehicle's mass — a loaded truck and a light scooter share the same limit on a given curve, because mass cancels out.

**In any circular motion the net force towards the centre must equal $\frac{m{v}^2}{r}$; in a vertical circle this force changes because gravity helps at the top and opposes at the bottom, and in a conical pendulum the horizontal component of the string's tension supplies it.

Vertical circle — a body on a string of length r:**

- At the top: $T+mg=\frac{m{v}^2}{r}$; at the bottom: $T-mg=\frac{m{v}^2}{r}$
- The string stays taut at the top only if $T\ge 0$, so the minimum speed there is $\sqrt{gr}$
- By conservation of energy, the minimum speed at the bottom is $\sqrt{5gr}$, where the tension is then $6mg$

Worked example 1. A 0.20 kg ball on a 0.80 m string is whirled in a vertical circle:

$$v_{\text{top}}=\sqrt{9.8\times 0.80}=2.8{\text{ m s}}^{-1}{v}_{\text{bottom}}=\sqrt{5\times 9.8\times 0.80}=6.26{\text{ m s}}^{-1}{T}_{\text{bottom}}=6\times 0.20\times 9.8=11.8\text{ N}$$

Conical pendulum. A bob on a string of length L moves in a horizontal circle, with the string at angle $\theta$ to the vertical:

- Vertically $T\cos \theta =mg$, and horizontally $T\sin \theta =\frac{m{v}^2}{r}$, so $\tan \theta =\frac{v^2}{rg}$
- With $r=L\sin \theta$, the time period is $t_p=2\pi \sqrt{\frac{L\cos \theta }{g}}$

Worked example 2. A 1.0 m conical pendulum swings with its string at 60° to the vertical:

$$t_p=2\pi \sqrt{\frac{1.0\times 0.5}{9.8}}=1.42\text{ s}$$

An everyday example. The chair swing ride at a Dussehra mela is a giant conical pendulum: as it spins faster, the chairs fly outwards to a larger angle.

The substance. No separate centrifugal force appears in these equations — seen from the ground, only tension, gravity and the normal reaction act, and their net inward part is the centripetal force.