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S1.4 Application: circular motion

[A] Uniform circular motion

Key Point 1.8

A particle moving in a circle of radius r at uniform speed v has an acceleration of magnitude v2/r, directed towards the centre of the circle (centripetal).

Commentary

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Commentary

  • speed is constant
  • but direction is changing
  • so velocity is changing
  • so there is an acceleration
  • it is a vector
  • we need to find its magnitude and its direction
 

Analysis

Analysis

Consider two points on the path, PQ

 

Graphic - No title - circularmotion

  • Construct vector triangle showing change in velocity.
  • Identify (average) acceleration:
    \[ \vec{a}_{av}= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{v}}{\Delta t}} \]
  • Direction of $\vec{a}_{av}$ is that of $\Delta \vec{v}$: towards centre
  • To find magnitude use
    \[ \mid \Delta \vec{v} \mid = \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{2 v \sin{\frac{\phi}{2}}}\]
    and
    \[ \Delta t= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{\mbox{\rm arc PQ}}{v} = \frac{r\phi}{v}} \]
    so that
    \[ \frac{\mid \Delta \vec{v} \mid}{\Delta t}= \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\frac{v^2}{r} \cdot \frac{2 \sin{\phi/2}}{\phi}} \]
  • As Δt approaches zero so does φ and
    \[ \sin{\frac{\phi}{2}} \xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{\simeq \phi/2} \]
  • Hence magnitude of acceleration is

    \[ a = \frac{v^2}{r} \cdot \frac{2 \times \phi/2}{\phi} = \frac{v^2}{r} \]

 

[B] Generalisations

Commentary

Commentary

  • Now $\vec{v}_P$ and $\vec{v}_{Q}$ differ in magnitude as well as direction.
  • So there is an extra contribution to $\Delta \vec{v}$
  • For Δt0 it is tangential to the path.

    Help?

    Help?

    This isn’t obvious from the figure. Let’s try to make it more plausible. As Δt becomes smaller so does φ. The vectors $\vec{v}_P$ and $\vec{v}_Q$ become closer to one another and (as drawn here) closer to the horizontal. The bit of the acceleration associated with the change of the speed (the ‘extra bit’ on the end of the $\vec{v}_Q$ vector) also comes closer to the horizontal …and this is tangential to the path (at the point where P and Q meet up).
     
  • Its magnitude is
    \[ \frac{\Delta v}{\Delta t} \rightarrow \frac{dv}{dt} \]
Graphic - No title - nonconstspeed
 
TVelocity and Acceleration of a Swinging Pendulum
APendulum vectors

Learning Resources

Textbook: HRW Chapter 4.7
Course Questions:
Self-Test Questions: