(455) Centripetal acceleration and force [IB Physics SL/HL]

Centripetal Acceleration and Force

Introduction

  • Centripetal means center-seeking.

  • Discusses the concept of circular motion and its implications in physics.

Acceleration

  • Definition: Acceleration is the change in velocity over time and is a vector (has magnitude and direction).

  • Acceleration can occur from changes in both speed and direction.

  • Example: A ball rolling in a straight line accelerates if its direction changes, even if its speed remains constant.

Concept Visualization

  • Visualize a ball on a flat surface.

  • If pushed in a straight line, it continues straight; tapping it laterally causes it to change direction towards the center, illustrating centripetal force.

  • Centripetal Force: Always directed toward the center of the circular path.

Centripetal Acceleration

  • Formula: ( a_c = \frac{v^2}{r} ) where:

    • ( a_c ): Centripetal acceleration (m/s²)

    • ( v ): Linear speed (m/s)

    • ( r ): Radius of the circle (m)

  • Both speed and radius must be considered for motion in a circle, reaffirming that direction change results in acceleration.

Centripetal Force

  • Derived from Newton's second law: ( F = ma )

  • Centripetal force formula: ( F = m \frac{v^2}{r} )

    • Where F is in Newtons, mass in kg, speed in m/s, and radius in meters.

    • Provides insight into the force required to maintain circular motion.

Angular & Linear Speed Relations

  • ( v = \omega r ) where ( \omega ): angular speed in radians/second.

  • Equivalent force formula: ( F = m \omega^2 r )

Example: Earth's Centripetal Acceleration

  • Radius of Earth: ( R = 6.4 \times 10^6 , m )

  • Calculate centripetal acceleration experienced due to Earth's rotation:

    1. Use ( A = \frac{v^2}{r} )

    2. Determine linear speed ( v ): Circumference of circle divided by the period of one rotation (24 hours).

    3. Convert 24 hours to seconds: 24 hours = 86,400 seconds.

    4. Substitute values to find acceleration: ( A = 4 \pi^2 \frac{R}{T^2} )

    5. Calculation yields: ( A \approx 0.0338 \text{ m/s²} )

Conclusion

  • The resulting acceleration is significantly less than standard gravitational acceleration, indicating a very small centripetal acceleration despite the Earth's rotation.

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