Forces, momentum and circular motion: study notes

1. Newton's laws and free-body diagrams

Learn the three laws in words; they are not in the booklet. First law: a body stays at rest or moves at constant velocity unless a resultant (net) force acts on it. Second law: the resultant force equals the rate of change of momentum, $F = \frac{\Delta p}{\Delta t}$, which for constant mass becomes $F = ma$. Third law: if body A exerts a force on body B, then B exerts a force on A that is equal in magnitude, opposite in direction and of the same type; the two forces act on different bodies, so they never cancel on a single free-body diagram. A free-body diagram shows one body only, as a dot or box, with every force on it drawn as an arrow from the body, labelled with its proper name: weight, normal reaction, tension, friction, drag, buoyancy. Do not draw the resultant, the acceleration or a centripetal force as separate arrows. Arrows should have lengths that reflect relative sizes where the question makes that possible (for example equal and opposite weight and normal reaction for a body at rest on a horizontal surface). Two marks are typically available: one for the complete set of forces with correct directions, one for correct labels.

2. Contact forces and field forces

Normal reaction $F_N$ acts perpendicular to the surface, away from it. Surface friction acts parallel to the surface and opposes relative motion or the tendency to move. Static friction adjusts itself up to a maximum, $F_f \le \mu_s F_N$ (booklet); slipping begins exactly when $F_f = \mu_s F_N$. Once sliding, dynamic friction is $F_f = \mu_d F_N$ (booklet), with $\mu_d < \mu_s$ usually. The elastic restoring force of a spring obeys Hooke's law $F_H = -kx$ (booklet), the minus sign showing it is directed towards the equilibrium position. Viscous drag on a sphere of radius $r$ moving slowly at speed $v$ through a fluid of viscosity $\eta$ is given by Stokes' law, $F_d = 6\pi \eta r v$ (booklet), opposing the velocity; terminal speed occurs when drag plus buoyancy balance weight. Buoyancy (Archimedes) is the upward force equal to the weight of fluid displaced, $F_b = \rho V g$ (booklet), where $\rho$ is the fluid density and $V$ the displaced volume. Weight is the field force $F_g = mg$ (booklet), acting vertically downward towards the centre of the Earth, with $g = 9.81\ \text{m s}^{-2}$. Tension acts along a string or rod, pulling away from the body at each end; a light inextensible string has the same tension throughout. On an incline, resolve weight into $mg\sin\theta$ down the slope and $mg\cos\theta$ into the slope, so that $F_N = mg\cos\theta$ when there is no other perpendicular force.

3. Linear momentum and impulse

Linear momentum is $p = mv$ (booklet), a vector in the direction of the velocity, unit kg m s$^{-1}$. Impulse is the change in momentum: $J = F\Delta t$ (booklet), and the identity $J = \Delta p$ is not printed in the booklet and must be learnt; the unit is N s, equivalent to kg m s$^{-1}$. On a force-time graph the impulse is the area between the line and the time axis, so a curved force-time graph is handled by counting squares or by approximating the shape as triangles and rectangles. Newton's second law in its general form, $F = \frac{\Delta p}{\Delta t}$ (booklet), is the one to use when the mass changes (a rocket ejecting fuel, sand falling onto a belt) or when the question gives momentum rather than acceleration. Because momentum is a vector, define a positive direction before substituting: a ball of mass $m$ hitting a wall at speed $v$ and rebounding at speed $v$ has $\Delta p = -2mv$ (initial direction taken as positive), not zero. A longer contact time for the same change in momentum means a smaller average force, which is why crumple zones and padded surfaces reduce injury.

4. Conservation of momentum, collisions and explosions

Learn the statement: the total momentum of a system is constant provided no resultant external force acts on it. This condition is not in the booklet and must be written down whenever you use conservation of momentum; the examiner usually looks for it. During a brief collision the internal forces between the bodies are large and the external forces (friction, air resistance) are negligible over the short time, so momentum is conserved even if friction acts before and after. In one dimension write total momentum before equals total momentum after, with signs for direction. In two dimensions apply the same equation separately to the $x$ and $y$ components; an explosion from rest, for example a stationary nucleus emitting a particle, gives fragments with equal and opposite momenta, and because $E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m}$ (booklet, listed under A.3) the lighter fragment carries the larger share of kinetic energy. An elastic collision conserves total kinetic energy as well as momentum; an inelastic collision does not, kinetic energy being transferred to internal energy, sound and deformation. A perfectly inelastic collision is one in which the bodies stick together and move with a common velocity; this is the case of greatest kinetic-energy loss consistent with momentum conservation. Momentum is conserved in every collision in which no resultant external force acts on the system (or the external impulse is negligible during the brief contact); kinetic energy is conserved only in elastic ones. Always finish a collision calculation by comparing kinetic energies before and after, stating whether the collision is elastic or inelastic and giving the energy lost.

5. Circular motion at constant speed

A body moving in a circle of radius $r$ at constant speed $v$ has a continuously changing velocity direction, so it is accelerating towards the centre and needs a resultant force towards the centre. The period $T$ is the time for one revolution, frequency $f = \frac{1}{T}$, and angular velocity is $\omega = \frac{2\pi}{T} = 2\pi f$ in rad s$^{-1}$. From the booklet: $v = \frac{2\pi r}{T} = \omega r$ and $a = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2}$, so the centripetal force is $F = \frac{mv^2}{r} = m\omega^2 r$. The essential rule to learn: the centripetal force is not a new force; it is the resultant of the real forces acting towards the centre. On a free-body diagram you draw only the real forces (tension, weight, normal reaction, friction, gravitational attraction) and then write the equation 'resultant towards centre $= \frac{mv^2}{r}$'. Because the force is always perpendicular to the velocity it does no work, so the speed stays constant while the velocity changes. Typical sources of centripetal force: tension for a whirled mass on a string, friction for a car on a flat bend, gravity for a satellite, the horizontal component of the normal reaction for a banked road, and the horizontal component of the tension for a conical pendulum.

6. Vertical circles, banked and horizontal turns

In a vertical circle the speed is not constant but the centripetal relation still holds at each point. For a mass on a string, at the lowest point tension and weight oppose, so $T - mg = \frac{mv^2}{r}$ and tension is greatest; at the highest point both act downward towards the centre, so $T + mg = \frac{mv^2}{r}$ and tension is least. The string stays taut at the top only if $v \ge \sqrt{gr}$; the minimum speed $v = \sqrt{gr}$ is the limiting case $T = 0$. For a car over a hump the same logic applies with normal reaction: $mg - F_N = \frac{mv^2}{r}$, and the car leaves the road when $F_N = 0$. On a horizontal (flat) turn friction supplies the whole centripetal force, so the maximum speed without skidding comes from $\mu_s F_N = \mu_s mg = \frac{mv^2}{r}$, giving $v_{max} = \sqrt{\mu_s g r}$, independent of mass. On a banked turn of angle $\theta$ with no friction, the normal reaction is tilted: vertically $F_N\cos\theta = mg$, horizontally $F_N\sin\theta = \frac{mv^2}{r}$, so $\tan\theta = \frac{v^2}{gr}$ at the design speed. Faster than this, friction must act down the slope; slower, friction acts up the slope. Writing the resolving equations in this two-line form typically earns the method marks even if a later arithmetic step goes wrong.

Formulas

Worked examples

Can you...

• State Newton's three laws in words, and explain why a third-law pair never appears on the same free-body diagram?
• Draw a complete, correctly labelled free-body diagram for a body on an incline, in a vertical circle and on a banked turn, with friction in the right direction?
• Resolve forces along and perpendicular to a slope and apply $F = ma$ to a single body or a connected system, using $F_N = mg\cos\theta$ and $F_f = \mu F_N$?
• Solve a one- or two-dimensional collision or explosion with conservation of momentum, state the no-external-force condition, and check kinetic energy to classify it as elastic or inelastic?
• Write 'resultant of real forces towards the centre $= \frac{mv^2}{r}$' for a car on a flat or banked bend, a mass on a string at the top and bottom of a vertical circle, and find a minimum or maximum speed condition?