Circular Motion

The complete acceleration vector has two perpendicular components:

1. Radial (centripetal) component: $a_r = -\omega^2r = -\frac{v^2}{r}$ (directed toward center)

2. Tangential component: $a_t = \frac{dv}{dt} = r\frac{d\omega}{dt} = r\alpha$ (tangent to circle)

Total acceleration magnitude: $a = \sqrt{a_r^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + \left(\frac{dv}{dt}\right)^2}$

Direction: Makes angle $\alpha$ with radius where: $\tan\alpha = \frac{a_t}{a_r} = \frac{dv/dt}{v^2/r}$

The radial component changes direction while tangential component changes speed.

• Centripetal force:

• Real force directed toward center of rotation
• Required for circular motion (provides necessary acceleration)
• Exists in inertial reference frames
• Examples: tension, friction, gravity, normal force
• Formula: $F_c = \frac{mv^2}{r} = m\omega^2r$

• Centrifugal force:

• Pseudo (fictitious) force appearing to act outward
• Only exists in rotating (non-inertial) reference frames
• Added to analysis to make Newton's laws applicable in rotating frames
• Formula: $F_{cf} = m\omega^2r$ (outward direction)
• Not a "new" force, just a mathematical construct

Use centripetal force when analyzing from stationary/inertial frames, and include centrifugal force only when analyzing from the perspective of the rotating system.

For a car on a banked track with angle $\theta$:

Without friction (ideal banking): $\tan\theta = \frac{v^2}{rg}$

With friction (maximum safe speed): $v_{max} = \sqrt{rg(\tan\theta + \mu_s)}$

Where:

• $v$ is speed
• $r$ is radius of turn
• $g$ is gravitational acceleration
• $\theta$ is banking angle
• $\mu_s$ is coefficient of static friction

Analysis:

• Normal force provides both vertical support and horizontal centripetal force
• Banking reduces or eliminates need for friction
• When speed exceeds ideal value, friction prevents outward skidding
• When speed is less than ideal, friction prevents inward skidding
• Maximum range of safe speeds: $\sqrt{rg(\tan\theta - \mu_s)} < v < \sqrt{rg(\tan\theta + \mu_s)}$

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

Force analysis: • Tension: $T$ along string • Gravity: $mg$ downward • Circular motion in horizontal plane of radius $r = L\sin\theta$

Vertical equilibrium: $T\cos\theta = mg$

Horizontal (centripetal force): $T\sin\theta = \frac{mv^2}{r} = m\omega^2r = m\omega^2L\sin\theta$

From these: $\tan\theta = \frac{\omega^2L\sin\theta}{g}$ $\omega^2 = \frac{g}{L\cos\theta}$

Period: $T = 2\pi\sqrt{\frac{L\cos\theta}{g}}$

Important relationships: • Larger angle $\theta$ → shorter period • Period depends on $\sqrt{L}$ (similar to simple pendulum) • Horizontal circle radius: $r = L\sin\theta$ • Vertical height: $h = L\cos\theta$

• 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.

The acceleration in circular motion decomposes into:

• Radial component: ar = -ω²r = -v²/r (points toward center, along -êr)

  • Represents the centripetal acceleration
  • Always present in circular motion, even at constant speed
  • Changes the direction of velocity without changing its magnitude

• Tangential component: at = dv/dt (points along êt)

  • Represents change in speed of the particle
  • Only present when speed is changing (non-uniform circular motion)
  • Changes the magnitude of velocity without changing its direction

Example: A car rounding a curve while accelerating experiences both components: centripetal acceleration keeping it on the circular path and tangential acceleration increasing its speed.

To prevent skidding on a horizontal circular turn:

The static friction force must be sufficient to provide the required centripetal force:

1. Required condition: fs ≥ mv²/r
2. Maximum available friction: fs(max) = μs·N = μs·mg

When these are combined:

• μs·mg ≥ mv²/r
• v² ≤ μs·g·r
• vmax = √(μs·g·r)

This means:

• Maximum speed increases with larger turn radius (r)
• Maximum speed increases with higher friction coefficient (μs)
• Speed must be reduced on wet or icy roads (lower μs)
• Skidding occurs when the centripetal force requirement exceeds the maximum available friction

Real-world application: Highway design standards specify minimum turn radii based on expected speeds and typical friction coefficients.

Inertial Reference Frame Analysis:

• Only real forces are included (gravity, normal force, friction, etc.)
• For an object moving in a circle: Σ F = ma, where a = ω²r toward center
• Requires a real centripetal force to maintain circular motion
• The resultant force points toward the center of rotation

Rotating Reference Frame Analysis:

• Includes both real forces and pseudo forces
• For the same object: must add centrifugal force (mω²r outward)
• Allows application of Newton's laws within rotating frame
• For stationary objects in rotating frame: forces sum to zero

Example: For a box at rest on a rotating platform, from an inertial frame, friction provides the centripetal force keeping it moving in a circle. From the rotating frame, friction balances the centrifugal force, resulting in equilibrium.

Calculation:

• Magnitude of required friction: f = mω²r
• f = (2 kg)(3 rad/s)²(0.5 m) = (2 kg)(9 rad²/s²)(0.5 m) = 9 N

Physical interpretation:

• In inertial frame: Friction provides the centripetal force (9 N inward) needed to keep the box moving in a circular path with proper acceleration
• In rotating frame: Friction (9 N inward) balances the centrifugal pseudo force (9 N outward), keeping the box stationary relative to the cabin

This demonstrates why a box appears stationary in a rotating reference frame despite actually undergoing circular motion - the frictional force provides exactly the acceleration needed to maintain circular motion when viewed from outside the rotating system.

The ratio of centrifugal acceleration at colatitude θ to centrifugal acceleration at the equator is sin(θ).

Derivation:

• Centrifugal acceleration = ω²r
• At any location with colatitude θ: r = R·sin(θ)
• At the equator (θ = 90°): r = R
• Ratio = (ω²·R·sin(θ))/(ω²·R) = sin(θ)

For example:

• At 60° colatitude (30° latitude), the centrifugal acceleration is sin(60°) = 0.866 times that at the equator
• At 30° colatitude (60° latitude), the centrifugal acceleration is sin(30°) = 0.5 times that at the equator

This explains why apparent weight variation due to Earth's rotation changes with latitude.