Q3.17 The Principle of Equivalence (H)

A car accelerates at 3 ms-2 along a straight road.
  1. A pendulum is suspended from the ceiling of the car. What angle does it make with the vertical? What is the direction of the deflection?
  2. A child, in the car, holds a balloon filled with helium, at the end of a string. What angle does it make with the vertical? What is the direction of the deflection?
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Hint

According to the Principle of Equivalence it is impossible to distinguish between the effects of an acceleration and the effects of a gravitational pull.
 

Solution

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Solution

  1. We can figure out what the deflection must be either by thinking from the perspective of a roadside observer, or from the perspective of a passenger in the car.

    First let’s think it through from the perspectives of the roadside observer.

    Let a=3 ms-2 denote the acceleration, of car (and, thus also the pendulum bob), and let T be the tension in the supporting string. The bob experiences two forces. The force exerted by the string, which has magnitude T and which acts along the string towards the point of support; and the bob weight, downwards.
    Graphic - No title - heliuma

    Applying Newton’s 2nd law to both vertical and horizontal directions we have

    \[ T\cos \theta -mg =0 \hspace{1cm}\mbox{\rm vertical direction}\hspace{1cm} \]
    and
    \[ T\sin \theta =m a \hspace{1cm}\mbox{\rm horizontal direction}\hspace{1cm} \]
    where we have resolved the string tension force into its horizontal and vertical components. Dividing the 2nd of these equations by the first we conclude that
    \[ \tan \theta = \frac{a}{g} = \frac{3}{9.8} \]
    implying θ=17 (2sf).

    We can also work things through from the perspectives of a car passenger. She sits in an accelerating reference frame and thus experiences a ‘fictitious’ force. We can write down that force explicitly (see S2.13) But it is helpful to invoke the Principle of Equivalence. This principle asserts that the effects on the physics experienced in a reference frame having an acceleration $\vec{a}_{NI}$ are indistinguishable from those due to an additional gravitational acceleration $\vec{g}_{NI} = -\vec{a}_{NI}$

    For the passenger, then, the effective total gravitational acceleration would be the sum of the true g (downwards) and this extra contribution.
    \[ \vec{g}_{\mbox{\rm eff}}= \vec{g}_{\mbox{\rm true}}+\vec{g}_{\mbox{\rm NI}} \]
    Graphic - No title - heliumb

    By inspection, the angle which $\vec{g}_{\mbox{\rm eff}}$ makes with the vertical satisfies

    \[ \tan \theta = \frac{g_{NI}}{g}= \frac{a}{g} \]
    the same angle we found above. From the perspectives of the car passenger the pendulum simply hangs in what, for her, is ‘down’.

  2. Having worked carefully through that argument we can handle the helium balloon easily.

    Helium balloons float ‘up’. So the string will again make an angle θ with the (true) vertical; moreover, the balloon will be deflected in the direction of the acceleration.
    Graphic - No title - heliumc

    You can reach this same –somewhat surprising– conclusion by thinking about the interplay between the tension force and the buoyancy force that the balloon experiences: the ‘real’ force to the right that the balloon needs to supply its acceleration (from the perspectives of the inertial frame) is then seen to come from the sideways push it gets from the air! But the Principle of Equivalence makes the argument a piece of cake. Perhaps, at this point you’ll think of Guideline 0.11.