9.3 Circular Motion, Gravitation & SHM

9.3 Circular Motion, Gravitation & SHM

Quick Answer: Circular motion MCQs hinge on $a_c = v^{2}/r$ and a real centripetal force provider; gravitation MCQs use $F = Gm_1m_2/r^{2}$ and $g = GM/R^{2}$; SHM MCQs use $a = -\omega^{2}x$ and $T = 2\pi/\omega$. Together they close the mechanics core for the PAF Aeronautical Engineering initial physics paper (FSc depth, ~30% academic weight).

Aircraft manoeuvre in curved paths, satellites orbit under gravity, and instruments oscillate—three reasons this trio appears on engineering entry tests more densely than on lighter GD Pilot academics.

Uniform Circular Motion (UCM)

Speed $v$ may be constant while velocity continuously changes direction. The acceleration is centripetal (toward centre):

$a_c = \frac{v^{2}}{r} = \omega^{2} r = \frac{4\pi^{2} r}{T^{2}}$

Relations:

$v = r\omega, \quad \omega = \frac{2\pi}{T} = 2\pi f$

Centripetal force magnitude:

$F_c = \frac{mv^{2}}{r} = m\omega^{2} r$

$F_c$ is not a new mysterious force; it is the required net radial force supplied by tension, gravity, friction, normal force, or aerodynamic lift.

Worked Example — Horizontal Circle

A $0.25,\text{kg}$ mass whirls on a $0.80,\text{m}$ string at $2.0,\text{rev/s}$. Find tension (neglect gravity for pure horizontal idealisation, or treat as conical separately—here assume horizontal support).

$\omega = 2\pi f = 4\pi\,\text{rad/s}$

$T = m\omega^{2} r = (0.25)(4\pi)^{2}(0.80) \approx 31.6\,\text{N}$

Banking (Conceptual Formula)

For a vehicle on a frictionless banked curve of angle $\theta$ and radius $r$,

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

Design speed $v = \sqrt{rg\tan\theta}$. With friction, a speed range exists around that design value—FSc stems usually stick to the frictionless or “maximum speed with $\mu$” variants.

Conical pendulum (related check): a bob of mass $m$ on string length $L$ moves in a horizontal circle of radius $r = L\sin\theta$. Vertical equilibrium gives $T\cos\theta = mg$; horizontal provides $T\sin\theta = mv^{2}/r$. Dividing yields $\tan\theta = v^{2}/(rg)$, the same trigonometric structure as banking—useful when an MCQ swaps the story but keeps the maths.

Vertical Circles

At the top of a vertical loop, both weight and normal may point toward centre:

$N_{\text{top}} + mg = \frac{mv^{2}}{r}$

Minimum speed at the top for just losing contact ($N = 0$): $v = \sqrt{gr}$. At the bottom, normal exceeds weight:

$N_{\text{bottom}} - mg = \frac{mv^{2}}{r}$

Gravitation

Newton’s law of universal gravitation:

$F = \frac{Gm_1 m_2}{r^{2}}$

$G \approx 6.67 \times 10^{-11},\text{N·m}^{2}\text{/kg}^{2}$. Direction: attractive along the line joining centres.

Gravitational field strength (acceleration due to gravity) at Earth’s surface:

$g = \frac{GM}{R^{2}}$

Variation with height $h$ above surface (for $h \ll R$ often approximated):

$g_h = \frac{GM}{(R+h)^{2}} \approx g\left(1 - \frac{2h}{R}\right)$

At depth $d$ inside a uniform sphere (FSc result):

$g_d = g\left(1 - \frac{d}{R}\right)$

Orbital Motion (Circular)

For satellite mass $m$ in circular orbit radius $r$:

$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r} \implies v = \sqrt{\frac{GM}{r}}$

Period from $T = 2\pi r / v$:

$T^{2} = \frac{4\pi^{2}}{GM} r^{3}$

(Kepler’s third law for circular orbits about a fixed central mass).

Escape speed from a planet’s surface:

$v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$

Worked Example — Compare $g$

If Earth’s radius were unchanged but mass doubled, $g$ would double because $g \propto M$. If mass unchanged but radius doubled, $g$ would fall to one-fourth because $g \propto 1/R^{2}$.

Simple Harmonic Motion (SHM)

A motion is SHM if the restoring acceleration is proportional to displacement and opposite in sign:

$a = -\omega^{2} x$

Standard solutions:

$x = A\cos(\omega t + \phi)$

$v = -A\omega\sin(\omega t + \phi), \quad v_{\max} = A\omega$

$a_{\max} = A\omega^{2}$

Mass–Spring

$\omega = \sqrt{\frac{k}{m}}, \quad T = 2\pi\sqrt{\frac{m}{k}}$

Total mechanical energy:

$E = \tfrac{1}{2}kA^{2} = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}kx^{2}$

At amplitude extremes, $v = 0$ and PE is maximum; at equilibrium, PE (elastic) is minimum and KE is maximum. Instantaneous speed at displacement $x$ follows from energy:

$v = \omega \sqrt{A^{2} - x^{2}}$

Simple Pendulum (Small Angles)

$T = 2\pi\sqrt{\frac{L}{g}}$

Independent of bob mass and (approximately) of amplitude when $\theta$ is small ($\sin\theta \approx \theta$ in radians). On the Moon, smaller $g$ means larger $T$.

Worked Example — Spring Period

Mass $0.50,\text{kg}$, spring constant $k = 200,\text{N/m}$:

$T = 2\pi\sqrt{\frac{0.50}{200}} = 2\pi\sqrt{0.0025} = 2\pi(0.05) = 0.314\,\text{s}$

$\omega = \sqrt{k/m} = \sqrt{400} = 20\,\text{rad/s}$

If amplitude $A = 0.04,\text{m}$, $v_{\max} = A\omega = 0.80,\text{m/s}$ and $E = \tfrac{1}{2}kA^{2} = 0.16,\text{J}$.

Worked Example — Pendulum

Length $1.0,\text{m}$, $g = 9.8,\text{m/s}^{2}$:

$T = 2\pi\sqrt{1/9.8} \approx 2.0\,\text{s}$

Bridging to Aeronautical Intuition (Exam-Safe)

You do not need aircraft structural codes for this paper, but linking ideas helps memory: turning flight requires a horizontal lift component (centripetal), orbital mechanics uses the same $v = \sqrt{GM/r}$ as textbook satellites, and vibration isolation is SHM energy trading. Keep answers algebraic and SI-clean.

Study Pace Reminder

Physics for AE initial is heavier than GD Pilot and commonly cited near 30% of the academic mix. Coaching-reported physics timings (~50 questions in ~25 minutes) are useful for mock drills but remain unofficial unless your cycle’s selection-centre brief or joinpaf.gov.pk states otherwise. Drill centripetal, $g$-variation, and SHM period formulas until they are automatic.

You now have the mechanics spine: units & kinematics → Newton/energy/momentum → circular/gravitation/SHM. Later physics chapters (fluids, thermo, EM, optics, modern) reuse the same SI and energy discipline.