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    Circular Motion

    Flashcards for topic Circular Motion

    Intermediate33 cardsGeneral

    Preview Cards

    Card 1

    Front

    What are the necessary conditions for a particle to move in circular motion? Explain in terms of forces, energy, and constraints.

    Back

    Necessary conditions for circular motion:

    Force requirement: • Net force toward center: F=mv2rF = \frac{mv^2}{r}F=rmv2​ • Must be perpendicular to velocity vector • Must be continuously maintained

    Energy considerations: • For uniform circular motion: kinetic energy remains constant • For non-uniform motion: work done by tangential force changes kinetic energy

    Constraints: • Physical constraint (like string, track, groove) OR • Field force (gravity, electromagnetic) providing centripetal force • Constraint must be able to withstand required tension/compression

    Additional requirements: • Initial velocity perpendicular to radius vector • Sufficient coefficient of friction (for motion on surfaces) • For vertical circles: minimum speed at top vmin=grv_{min} = \sqrt{gr}vmin​=gr​

    Without these conditions, the particle would follow a straight-line path (Newton's 1st law).

    Card 2

    Front

    When a Coriolis force is needed in addition to centrifugal force in rotating reference frames, and how is it calculated?

    Back

    The Coriolis force is needed when: • Working in a rotating reference frame • AND the particle is moving relative to that rotating frame • (Centrifugal force alone is sufficient only for stationary objects in rotating frames)

    Calculation: F⃗Coriolis=−2m(ω⃗×v⃗rel)\vec{F}_{Coriolis} = -2m(\vec{\omega} \times \vec{v}_{rel})FCoriolis​=−2m(ω×vrel​)

    Where: • mmm is particle mass • ω⃗\vec{\omega}ω is angular velocity vector of rotating frame • v⃗rel\vec{v}_{rel}vrel​ is velocity of particle relative to rotating frame

    Properties: • Always perpendicular to both rotation axis and relative velocity • Magnitude: FCoriolis=2mωvrelsin⁡ϕF_{Coriolis} = 2m\omega v_{rel}\sin\phiFCoriolis​=2mωvrel​sinϕ (where ϕ\phiϕ is angle between vectors) • Causes apparent deflection of moving objects in rotating frames • Examples: hurricane circulation, apparent paths of projectiles on Earth

    The combination of centrifugal and Coriolis forces allows application of Newton's laws in non-inertial rotating reference frames.

    Card 3

    Front

    A conical pendulum makes a fixed angle with the vertical. Derive the relationship between the angle, length, angular speed, and period.

    Back

    For a conical pendulum (mass on string making angle θ\thetaθ with vertical while rotating):

    Force analysis: • Tension: TTT along string • Gravity: mgmgmg downward • Circular motion in horizontal plane of radius r=Lsin⁡θr = L\sin\thetar=Lsinθ

    Vertical equilibrium: Tcos⁡θ=mgT\cos\theta = mgTcosθ=mg

    Horizontal (centripetal force): Tsin⁡θ=mv2r=mω2r=mω2Lsin⁡θT\sin\theta = \frac{mv^2}{r} = m\omega^2r = m\omega^2L\sin\thetaTsinθ=rmv2​=mω2r=mω2Lsinθ

    From these: tan⁡θ=ω2Lsin⁡θg\tan\theta = \frac{\omega^2L\sin\theta}{g}tanθ=gω2Lsinθ​ ω2=gLcos⁡θ\omega^2 = \frac{g}{L\cos\theta}ω2=Lcosθg​

    Period: T=2πLcos⁡θgT = 2\pi\sqrt{\frac{L\cos\theta}{g}}T=2πgLcosθ​​

    Important relationships: • Larger angle θ\thetaθ → shorter period • Period depends on L\sqrt{L}L​ (similar to simple pendulum) • Horizontal circle radius: r=Lsin⁡θr = L\sin\thetar=Lsinθ • Vertical height: h=Lcos⁡θh = L\cos\thetah=Lcosθ

    Card 4

    Front

    What are the unit vectors êr and êt in circular motion, and how are they defined in relation to a particle moving in a circle?

    Back

    • Radial unit vector (êr): Points outward along the radius from the center O to the particle P
    • Tangential unit vector (êt): Points perpendicular to êr in the direction of increasing θ (tangent to the circle)

    These unit vectors form a local coordinate system that moves with the particle, providing a natural basis for expressing forces and motion in circular paths.

    Note: Unlike fixed Cartesian unit vectors (î, ĵ), these unit vectors continuously change direction as the particle moves around the circle.

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    Card 5

    Front

    Express the radial unit vector êr and the tangential unit vector êt in terms of Cartesian unit vectors î and ĵ for a particle at angular position θ.

    Back

    • Radial unit vector: êr = î cos θ + ĵ sin θ
    • Tangential unit vector: êt = -î sin θ + ĵ cos θ

    These expressions show how the circular motion unit vectors can be decomposed into components along the fixed coordinate axes, essential for analyzing motion and forces in rotating systems.

    Example: At θ = 45°, êr = (î + ĵ)/√2 and êt = (-î + ĵ)/√2

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    Card 6

    Front

    In a physical scenario involving nonuniform circular motion, what determines whether the radial or tangential component of acceleration will dominate, and how does this affect the motion?

    Back

    The relative dominance of acceleration components in nonuniform circular motion:

    Radial component (v²/r) dominates when:

    • The circular path has a small radius
    • The speed is very high
    • The rate of speed change is small
    • Example: A car taking a sharp turn at high speed (primary concern is preventing skidding outward)

    Tangential component (dv/dt) dominates when:

    • There is rapid acceleration or deceleration
    • The radius is large
    • The speed is relatively low
    • Example: A train starting from rest on a curved track (primarily feels forward thrust)

    Effects on motion:

    • Radial-dominated: Object maintains its circular path but experiences strong inward forces
    • Tangential-dominated: Object changes speed significantly while following the path
    • Balanced components: Object follows a spiral-like path if unconstrained

    Practical application: Vehicle designers must account for both components when designing suspension systems for vehicles that accelerate through turns.

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    Card 7

    Front

    For a vehicle on a banked curve with banking angle θ, what is the "ideal speed" formula, and what happens when a vehicle travels at exactly this speed?

    Back

    • The ideal speed formula is: v = √(rg·tanθ) Where:

      • v = velocity
      • r = radius of the curve
      • g = acceleration due to gravity
      • θ = banking angle
    • When a vehicle travels at exactly this speed:

      • No friction is needed to maintain the circular path
      • The normal force alone provides the necessary centripetal force
      • The vehicle neither slides up nor down the banking
      • The horizontal component of the normal force exactly equals mv²/r
      • The vertical component of the normal force exactly equals the weight mg
    • This represents the perfect equilibrium condition for the banked turn

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    Card 8

    Front

    How does the apparent vertical direction differ from true vertical on Earth's surface, and what practical implications does this have?

    Back

    • True vertical points toward Earth's center (direction of gravitational force mg)
    • Apparent vertical deviates by angle α, where:
      • tanα = (ω²R·sinθ·cosθ)/(g - ω²R·sin²θ)
      • α is maximum at mid-latitudes
      • α = 0 at poles and equator
    • Practical implications:
      • Plumb lines align with apparent vertical, not true vertical
      • Buildings are constructed parallel to apparent vertical
      • Water surfaces at rest are perpendicular to apparent vertical
      • This deviation must be accounted for in precise surveying and astronomical measurements
      • Navigation systems must correct for this difference in high-precision applications
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    Card 9

    Front

    At what latitude on Earth would the reduction in apparent weight be exactly half of its maximum possible value, and why is this significant?

    Back

    • Maximum weight reduction occurs at equator (θ = 90°): mg' = mg - mω²R
    • At latitude λ (colatitude θ = 90° - λ), the reduction factor depends on sin²θ
    • For half the maximum reduction: sin²θ = 0.5
    • Therefore: sin²(90° - λ) = 0.5
    • Solving: λ = 45° (occurs at 45° latitude north and south)
    • Significance:
      • This demonstrates that apparent weight reduction is not linearly related to latitude
      • The effect follows a sin² curve, decreasing more rapidly near equator
      • At 45° latitude, objects weigh halfway between their equatorial and polar weights
      • This has implications for precise scientific measurements that depend on weight or the local value of g
      • Geological surveys and gravitational models must account for this variation
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    Card 10

    Front

    What is colatitude, and how does it relate to a point's circular motion due to Earth's rotation?

    Back

    Colatitude is the angle (θ) between Earth's rotation axis and the radius vector from Earth's center to a specific point on its surface.

    Key properties:

    • Complement to the geographical latitude (90° - latitude)
    • Determines the radius of the circular path a point follows during Earth's rotation
    • The radius of this circular path = R·sin(θ), where R is Earth's radius

    For example, at the equator, colatitude is 90° and points move in a circle of radius equal to Earth's radius. At the poles, colatitude is 0°, and points remain stationary with respect to Earth's rotation.

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