IB DP Physics: HL复习笔记6.1.3 Centripetal Acceleration

Centripetal Acceleration

  • For an object moving in a circle:
    • The acceleration is towards the centre of the circle
    • The magnitude of the centripetal acceleration a is:

3.-Calculating-Centripetal-Acceleration-equation-1-1

  • Where:
    • a = centripetal acceleration (m s–2)
    • v = linear speed (m s–1)
    • r = radius of orbit (m)
  • Uniform circular motion is continuously changing direction, and therefore is constantly changing velocity
    • The object must therefore be accelerating
  • This is called the centripetal acceleration

Direction of the Centripetal Acceleration

  • The centripetal acceleration is perpendicular to the direction of the linear velocity
    • Centripetal means it acts towards the centre of the circular path

6-1-3-direction-of-acceleration_sl-physics-rn

Slide a ruler parallel to Δv towards the circle. Midway between A and B, Δv points towards the centre of the circle. This is the same direction as the centripetal acceleration

  • If an object moves through a section of a circle during some time Δt
  • The change in velocity during this time is Δv
  • The centripetal acceleration is Δv (a vector) divided by Δt (a scalar)
    • The centripetal acceleration points in the same direction as the change in velocity Δv
  • The centripetal acceleration is caused by a centripetal force of constant magnitude that also acts perpendicular to the direction of motion (towards the centre)
  • There is no component of the centripetal force in the direction of the velocity
    • Therefore, there is no acceleration in the direction of the velocity
    • Hence, there is uniform motion at constant speed
  • Therefore, the centripetal acceleration and force act in the same direction

Magnitude of the Centripetal Acceleration

  • In the diagram above notice how the angle Δθ is defined in terms of the arc length vΔt and the radius r
  • v is the magnitude of v1 and v2
  • Notes for deriving the equation for centripetal acceleration:
    • The vector triangle should be formed so that Δv is horizontal
    • The velocity vectors v should be of the same length
    • Hence, the vertical line bisects the angle Δθ and the vector Δv
    • Use trigonometry for one of the small triangles
    • The small-angle approximation requires that the angles are in radians
    • The two equations for Δθ lead to the magnitude of the centripetal acceleration

6-1-3-magnitude-of-acceleration_sl-physics-rn

Deriving the equation for the magnitude of the centripetal acceleration

  • This leads to the equation for centripetal acceleration:

3.-Calculating-Centripetal-Acceleration-equation-1-1

  • Using the equation relating angular speed ω and linear speed v:

 v = r⍵

  • These equations can be combined to give another form of the centripetal acceleration equation:

3.-Calculating-Centripetal-Acceleration-equation-2

  • Where:
    • a = centripetal acceleration (m s2)
    • v = linear speed (m s1)
    • ⍵ = angular speed (rad s1)
    • r = radius of the orbit (m)
  • Uniform centripetal acceleration is defined as:

The acceleration of an object towards the centre of a circle when an object is in motion (rotating) around a circle at a constant speed12.2.1.3-Centripetal-acceleration-diagram_1

Centripetal acceleration is always directed toward the centre of the circle and is perpendicular to the object’s velocity

Worked Example

A domestic washing machine has a spin cycle of 1200 rpm (revolutions per minute) and a diameter of 50 cm.

Calculate the centripetal acceleration experienced by the washing during the spin cycle.

Step 1: List the known quantities

    • Radius of the drum, r = ½ × 50 cm = 25 cm

Step 2: Convert the revolutions per minute to revolutions per second

1200 ÷ 60 = 20 rev s−1

Step 3: Convert revolutions per second to angular speed in radians per second

1 rev s–1 = 2π rad s–1

20 rev s–1 = 40π rad s–1 = ω

Step 4: Write the equation linking centripetal acceleration and angular speed

a = 2

Step 5: Calculate the centripetal acceleration

a = (25 × 10−2) × (40π)2

Step 6: State the final answer

a = 3900 m s−2 (2 s.f.)

Worked Example

A ball tied to a string is rotating in a horizontal circle with a radius of 1.5 m and an angular speed of 3.5 rad s−1.

Calculate its centripetal acceleration if the radius was twice as large and angular speed was twice as fast.

WE-Centripetal-acceleration-answer-image

Exam Tip

The key takeaways for an object moving in a circle are:

  • The magnitude of the velocity vector does not change
  • The direction of the velocity vector does change
  • Therefore, there is an acceleration despite the speed not changing

 

 

 

 

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