Advanced Mechanics

Module 5 is worth about a quarter of the HSC paper. Motion in gravitational fields is the single most heavily examined inquiry question in Year 12 Physics (about 12 marks a year on average), so give it the most time.

• Projectile motion: split motion into independent horizontal (constant velocity) and vertical (constant acceleration $g$) components; solve for time of flight, range, maximum height and final velocity; describe how air resistance changes the path.
• Circular motion: uniform circular motion, centripetal acceleration and force, angular velocity, what provides the centripetal force in different situations, banked tracks, work and energy in circular motion, torque.
• Motion in gravitational fields: Newton's law of universal gravitation, gravitational field strength, orbits and Kepler's laws, gravitational potential energy, total energy of a satellite, escape velocity.

Throughout, take $g=9.8{\text{m s}}^{-2}$ near Earth's surface unless the question says otherwise.

A projectile is any object that, after launch, moves only under the influence of gravity. With air resistance ignored, the only force is the object's weight, which acts vertically downwards. That single idea drives everything else:

• Horizontally there is no force, so horizontal velocity is constant: $v_x=u_x$.
• Vertically the acceleration is constant: $a_y=9.8{\text{m s}}^{-2}$ downwards, at every point of the flight, including the top.
• The two components are independent. They are linked only by time, which is the same for both.
• Constant horizontal velocity plus uniformly accelerated vertical motion gives a parabolic path.

Formulas (data sheet)

Resolve the launch velocity $u$ at angle $\theta$ to the horizontal into $u_x=u\cos \theta$ and $u_y=u\sin \theta$. Then use the constant-acceleration equations on each component:

Here $s$ is displacement (m), $u$ and $v$ are initial and final velocity (${\text{m s}}^{-1}$), $a$ is acceleration (${\text{m s}}^{-2}$) and $t$ is time (s). Horizontally $a=0$, so $s_x=u_{x}t$. Vertically, choose a sign convention (for example, up is positive, so $a_y=-9.8{\text{m s}}^{-2}$) and stick to it.

Useful facts: at maximum height $v_y=0$; on level ground the flight is symmetrical; when launched from a height it is not, so use the full quadratic for time.

Worked example

A ball is launched at $22{\text{m s}}^{-1}$ at $50^{\circ }$ above the horizontal from the edge of a cliff 15 m high. Find the time of flight, range and impact velocity. (Up is positive, $g=9.8{\text{m s}}^{-2}$. Keep unrounded values in your calculator; the numbers shown are rounded.)

Step 1: components.

Step 2: time of flight. The ball lands 15 m below the launch point, so $s_y=-15$ m:

Rearrange to $4.9t^2-16.85t-15=0$ and take the positive root: $t=4.17$ s.

Step 3: range.

Step 4: impact velocity. The final vertical velocity is

(downwards). Combine the components:

So the ball hits the ground at $27.9{\text{m s}}^{-1}$, $59.5^{\circ }$ below the horizontal.

The maximum height above the cliff top is $u_y^2/2g=14.5$ m. Notice that a velocity answer needs a magnitude and a direction.

Air resistance

The syllabus assumes zero air resistance, but questions sometimes ask you to describe its effect qualitatively. Drag acts opposite to the velocity, so it reduces both components. The projectile reaches a lower maximum height, has a shorter range, and the path is no longer symmetrical: it comes down more steeply than it went up. A full answer names the force, its direction and its effect on each component.

Practical: projectile launcher

A common practical investigation is a ball rolled off a ramp or fired from a launcher, with range measured for different launch heights, speeds or angles. Know how to improve validity (control launch speed, measure launch height to the centre of the ball), reliability (repeat trials and average) and accuracy (video analysis rather than eyeballing landing spots). Expect to compare measured range with the predicted range and explain the difference using air resistance and measurement uncertainty.

Uniform circular motion is motion in a circle at constant speed. The velocity is always tangent to the circle, so its direction keeps changing. A changing velocity means the object is accelerating, and Newton's second law says a net force must cause it. That net force points towards the centre and is called the centripetal force.

The centripetal force is not a new kind of force. It is the name for whichever real force (or component) points to the centre: friction on a flat road, tension in a string, a component of the normal force on a banked track, or gravity for an orbiting satellite. "Centrifugal force" is not a real force in an inertial frame; do not draw it on a force diagram.

Formulas (data sheet)

• $a_c$: centripetal acceleration (${\text{m s}}^{-2}$), directed towards the centre.
• $F_c$: centripetal (net) force (N); $m$ in kg; $r$ is the radius (m).
• $T$: period, the time for one revolution (s). Frequency $f=1/T$.
• $\omega$: angular velocity (${\text{rad s}}^{-1}$). For a full revolution, $\omega =2\pi /T$, and the linear speed is $v=\omega r$.

Worked example: whirling a ball

A 0.25 kg ball on a 0.60 m string is swung in a horizontal circle at 2.5 revolutions per second (treat the string as horizontal).

The tension in the string provides this 37.0 N centripetal force.

Banked tracks

On a banked road angled at $\theta$, the normal force is perpendicular to the road surface, so it has a horizontal component towards the centre of the curve. At the design speed, that component alone provides the centripetal force and no friction is needed. Resolving forces gives $N\sin \theta =m{v}^2/r$ and $N\cos \theta =mg$, so

The mass cancels, so the design speed is the same for every vehicle. Above the design speed, friction acts down the slope to add extra centripetal force; below it, friction acts up the slope.

Example. A curve of radius 150 m is to be banked so that cars at $25{\text{m s}}^{-1}$ need no friction: $\tan \theta =25^2/(150\times 9.8)$, so $\theta =23.0^{\circ }$.

Energy, work and torque

In uniform circular motion the centripetal force is always perpendicular to the velocity, so it does no work ($W=Fs\cos 90^{\circ }=0$). The speed and kinetic energy stay constant; only the direction changes.

Torque is the turning effect of a force about an axis:

where $r$ is the distance from the pivot to where the force acts (m), $\theta$ is the angle between $r$ and $F$, and torque is in N m. Torque is largest when the force is perpendicular to the lever arm. This links straight to the torque on a motor coil in Module 6.

Practical: centripetal force

In the whirling-stopper investigation, a hanging mass provides the tension (centripetal force) for a stopper swung in a horizontal circle. Plotting $F_c$ against $v^2$ (or $T^{-2}$) should give a straight line through the origin, confirming $F_c\propto v^2$ for constant $m$ and $r$.

Newton's law of universal gravitation

$F=GMm/r^2$ is on the data sheet. Dividing by the test mass $m$ gives the field strength $g$.

• $G=6.67\times 10^{-11}{\text{N m}}^2{\text{kg}}^{-2}$; $M$ and $m$ in kg.
• $r$ is the distance between the centres of mass (m). For a satellite, $r$ = planet radius + altitude.
• $g$ is the gravitational field strength (${\text{N kg}}^{-1}$, which equals ${\text{m s}}^{-2}$). It is a vector field pointing towards the centre of the mass.

The force is an inverse-square law: doubling $r$ quarters $F$ and $g$. Field diagrams show radial lines pointing inwards, closer together where the field is stronger.

Orbits and Kepler's laws

For a circular orbit, gravity provides the centripetal force. Setting $GMm/r^2=m{v}^2/r$ gives the orbital speed, and using $v=2\pi r/T$ gives Kepler's third law:

The satellite's own mass cancels, so orbital speed and period depend only on the central mass and the radius. A higher orbit is slower and has a longer period.

Kepler's three laws:

1. Planets move in ellipses with the Sun at one focus.
2. A line from the Sun to a planet sweeps out equal areas in equal times, so the planet moves faster when it is closer to the Sun.
3. $r^3/T^2$ is the same for every body orbiting the same central mass.

In an elliptical orbit the total energy $K+U$ stays constant: closer to the Sun, $U$ is more negative so $K$ is larger and the planet moves faster. For an ellipse, $r$ in Kepler's third law is the semi-major axis.

Low Earth orbits (a few hundred km up) have periods of about 90 minutes and suit imaging, the space station and some communications. Geostationary orbits sit about 36 000 km above the equator with a 24-hour period, so the satellite stays above one point on the ground (used for weather and communications).

Energy in a gravitational field

$U$ is on the data sheet. Be able to derive $v_{esc}$ from $K+U=0$ whether or not it is listed.

$U$ is gravitational potential energy (J), defined as zero at infinite separation, which is why it is negative. It is the work done to bring the mass from infinity to $r$. The work needed to move a mass between two points is $\Delta U=U_{final}-U_{initial}$.

For a circular orbit, $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$, so the total energy is

Moving to a higher orbit makes $E$ less negative, so energy must be added, even though the satellite ends up moving more slowly. Conversely, atmospheric drag removes energy, the satellite drops to a lower orbit and actually speeds up. Escape velocity is the minimum launch speed (ignoring air resistance) for which $K+U\ge 0$; it does not depend on the mass being launched.

Worked example: a stationary orbit around Mars

Find the radius and speed of an orbit that keeps a 300 kg satellite above one point on Mars, and its total energy. Mars: $M=6.42\times 10^{23}$ kg, radius $3.39\times 10^6$ m, rotation period $88642$ s.

Radius, from Kepler's third law:

which gives $r=2.04\times 10^7$ m. Subtracting the radius of Mars, the altitude is $1.70\times 10^7$ m, about 17 000 km.

Speed: $v=\sqrt{GM/r}=1.45\times 10^3{\text{m s}}^{-1}$. (Check: $2\pi r/T$ gives the same.)

Total energy (using the unrounded $r$):

so $E=-3.14\times 10^8$ J.

Band 6 depth

Top answers derive relationships rather than just quote them (for example, equate $F_g$ and $F_c$ to explain why mass cancels), state where $r$ is measured from, and explain energy changes in terms of work done. When comparing orbits, comment on speed, period, kinetic, potential and total energy together.

• Using the altitude instead of the orbital radius in $r$ (add the planet's radius).
• Saying acceleration is zero at the top of a projectile's path. Only the vertical velocity is zero.
• Using the symmetrical-flight shortcut when the launch and landing heights differ.
• Mixing degrees and radians, or forgetting to convert revolutions per minute to ${\text{rad s}}^{-1}$ ($1\text{rev}=2\pi \text{rad}$).
• Drawing a "centrifugal force" outwards, or adding the centripetal force as an extra force on a free-body diagram.
• Forgetting the negative sign in $U=-GMm/r$, or thinking a higher orbit has less total energy.
• Giving a velocity without a direction.

• Calculate: show the formula, substitution with units, and the answer to sensible significant figures (match the data, usually 2 or 3). Marks are given for working, so write it even when you can do it in your head.
• Explain: give the cause and the effect and link them with "because" or "so". For projectiles that means naming gravity as the only force and saying what it does to each component.
• Analyse: identify the components or factors and show how they relate, often with a calculation or a graph.
• Evaluate / assess: make a judgement backed by evidence, such as whether an experimental result supports a prediction and how big the uncertainty is.
• Use the data sheet for constants and formulas, but know what each symbol means and when the formula applies (for example $v=\sqrt{GM/r}$ is for circular orbits only).
• For derivations (such as Kepler's third law or the banked-track equation), start from the physics ($F_g=F_c$, or resolving forces) and show each algebraic step.
• Sketch a labelled diagram for projectile and circular-motion problems; markers reward clear sign conventions.

• Module 6: charged particles between parallel plates follow the same parabolic path as projectiles, and charged particles in magnetic fields move in circles ($F_c=qvB$). Torque reappears in the DC motor.
• Module 7: relativity changes how time and length are measured for fast-moving satellites (GPS clocks).
• Module 8: Kepler's laws and orbital motion underpin the Bohr model's picture of electrons orbiting a nucleus, and Coulomb's law (Year 11) has the same inverse-square form and governs Rutherford scattering.
• Year 11 Dynamics: Newton's laws, work and energy are assumed knowledge here.