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Lesson 6 · 45 min

Gravitation, circular motion and oscillations

Use inward acceleration, gravitational fields and restoring forces to explain curved and periodic motion.

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Gravitation, circular motion and oscillations

Constant with changing

includes direction as well as . A body moving at constant around a circle changes direction continuously and therefore accelerates. Uniform circular points toward the centre with magnitude. The required resultant inward is mv²/r. Centripetal is the role played by net inward , not a new added to tension, or gravity. Identify the actual provider: tension in a whirled string, gravity in an ideal orbit or in a level-road turn. If the net inward vanishes, the existing tangential does not automatically point radially outward.

Key equation
a=v²r

Period and angular

Period T is one-revolution time and f revolutions per second, so. Circumference2πr divided by period gives. One revolution is2π radians; angular ω and. Circular can therefore beω²r. Points on a rigid rotating wheel share angular but a larger-radius point has greater linear . Angular is rate of change of angular in a fixed-axis simple model. Convert revolutions per minute to revolutions per second before using SI time. State whether a quantity describes angular change or linear path length.

Key equation
f=1T
v=2πrT
=2πT
v=rω

Horizontal and vertical diagrams

On a horizontal frictionless table tension can all inward . In a vertical circle, gravity contributes differently at the top and bottom: at the top weight and tension can both point inward, while at the bottom tension points inward and weight outward. Tension is therefore not automatically constant. For a conical pendulum, vertical tension component balances weight and horizontal component supplies centripetal . The same tension is resolved into components; it is not three independent forces. An apparent centrifugal is used in a rotating reference frame, not added as an extra outward physical interaction when applying an inertial-frame diagram.

Universal gravitation and field strength

Two point masses, or suitable spherically symmetric bodies outside their surfaces, attract with, where r is centre-to-centre separation. Doubling r quarters at fixed masses. Gravitational field strength is per test mass in N/kg, equivalent to m/s². A test mass cancels when gravity equals ma, explaining mass-independent ideal free fall. Near Earth's surface g changes little over small heights, giving the constant-g approximation. Across the surface, , Earth's rotation and its nonspherical shape contribute to variation; at significant altitude the increased centre distance also reduces the gravitational field contribution.

Key equation
F=GMmr²
g=GMr²

Gravitational potential and conservative work

Near the surface small height changes useΔU. For large separations from an isolated spherical mass, with zero at infinity. Negative potential indicates a bound attractive system under that reference. Raising an object makes its potential less negative. Gravitational potential is per unit test mass, not field strength. Gravity is conservative: work between two positions depends on endpoints rather than path and the net work around a closed path is zero. is generally nonconservative and transfers mechanical to internal stores. Escape √(2GM/r) follows the ideal no- account and does not depend on launched mass.

Key equation
=mgΔh
U=−GMmr

Circular orbits and weightlessness

Equating gravity to centripetal gives and. Orbital radius includes planet radius plus altitude. The period is2π√(r³/GM), so a larger circular orbit has a longer period and smaller . Orbiting astronauts are still affected by gravity; they and the craft fall together, reducing ordinary support forces and producing apparent weightlessness. A geostationary orbit must have the appropriate Earth-rotation period, lie in the equatorial plane and follow Earth's rotation direction. A satellite with the same period but the wrong plane is not stationary over one point in the same sense.

Key equation
GMmr²=mv²r
v=√(GM/r)

Simple harmonic motion

SHM requires proportional to from and directed toward it. The restoring minus sign matters. is maximum at and zero at extreme displacements; magnitude is greatest at the extremes. A mass on an ideal spring has period. A simple small-angle pendulum has approximate period, with effective length measured from pivot to bob centre. A large swing violates the small-angle approximation. Amplitude is the maximum and differs from period and . Do not apply the spring formula to every periodic system merely because it moves back and forth.

Key equation
T=2π√(m/k)
T=2π√(L/g)
a=−ω²x
Where: T: period (s); m: mass on the spring (kg); k: spring constant (N/m). T: period (s); L: length from pivot to bob centre (m); g: acceleration of free fall (m/s²). a: acceleration (m/s²); ω: angular frequency (rad/s); x: displacement from equilibrium (m). The minus sign shows acceleration points back toward equilibrium.

, damping and resonance

Ideal oscillations exchange kinetic and potential while total mechanical remains constant. An ideal spring's total oscillation is½kA² for amplitude A. At kinetic is greatest; at extremes it is zero and potential greatest. Damping transfers away and reduces amplitude. A periodic driver can replace losses, and a driving near natural can create resonance with a large response depending on damping. The driver supplies that ; resonance does not create it. A bridge or instrument example should explain a driving mechanism and natural response rather than treating every vibration as necessarily resonant.

Testing a pendulum model

Measure several complete oscillations and divide total time by their number to reduce fractional reaction-time effects. Use small amplitude and a consistent crossing point, repeat measurements and vary length in a suitable range. The ideal relation gives a straight T²-versus-L graph with gradient4π²/g. A T-versus-L graph is curved. Measure to the bob's centre of mass, not its upper surface. Avoid pushing it into unwanted elliptical motion. A graph intercept or systematic discrepancy can reveal a length offset or other model limitation; it should not be dismissed as generic human error without explanation.

Key equation
T²=(4π²/g)L

Self-check and proportional reasoning

A0.5 kg object at4 m/s in radius2 m needs inward force0.5 ×16/2 =4 N. If its string breaks, it initially follows a tangent subject to remaining forces. Doubling gravitational separation quarters attraction. Multiplying a small-angle pendulum length by four doubles its period at fixed g. These are different proportional laws: v²/r,1/r² and√L respectively. State which variables are held fixed in each comparison. In numerical work retain consistent SI units and interpret signs from an explicit frame. A negative gravitational potential is compatible with a positive escape- requirement because must be supplied to reach the zero-at-infinity reference.

Topic revision bank

Key rules & principles

  • Centripetal means resultant inward .
  • Orbit radius is measured from the central body centre.
  • The pendulum approximation needs small amplitude.

Worked examples

Find force for0.2 kg moving6 m/s around radius0.9 m.
  1. Use mv²/r.
  2. Substitute0.2 ×36/0.9.
  3. The result is inward.

Answer: 8 N.

A1.0 m pendulum makes20 oscillations in40 s. Estimate g.
  1. T =40/20 =2.0 s.
  2. g =4π²L/T².
  3. Use4π² ×1/4.

Answer: Approximately9.87 m/s².

Related videos and resources

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Newtonian Gravity: Crash Course Physics #8

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Newtonian gravitational concepts; paired with circular motion.

Uniform Circular Motion: Crash Course Physics #7

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Centripetal acceleration and force; introductory examples use an international context.

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