Q2.6 A consolidation exercise (S)

Establish from first principles the acceleration of a particle moving at constant speed v in a circle of radius r.
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This is an invitation to rework the argument which you should find in your lecture notes.
 

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Solution

Graphic - No title - circularmotion

Consider two points on the circular path, P and Q say (above, left) It helps to take them to be not too far apart (since we are going to consider the limit in which they effectively coincide).

Now draw a triangle (right) in which two sides are formed by vectors giving $\vec{v}_P$ (the velocity at P) and $\vec{v}_Q$ (the velocity at Q), tail-to-tail. Then the third side of the triangle automatically gives the change in the velocity as the particle moves from P to Q:

\[ \Delta \vec{v} =\vec{v}_P - \vec{v}_Q \]

The average acceleration over this interval (of duration Δt, say) is then (recall Equation 1.4, written for the case of 1D motion)

\[ \vec{a}_{\mbox{\rm av}} = \frac{\Delta \vec{v}}{\Delta t} \]
The instantaneous acceleration at the single point picked out when we think of Δt being small, so that P and Q coincide, is:
\[ \vec{a}= \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{v}}{\Delta t} \]

By inspection we can see that the direction of $\vec{a}$ is that of $\vec{a}_{\mbox{\rm av}}$, and thus of $\Delta \vec{v}$ itself. It is towards the centre of the circle (it happens to be downwards, because of our choice of P and Q).

To find magnitude of $\vec{a}$ we use a little trigonometry on the vector triangle, to give

Δv=2vsinφ/2
Moreover since v is constant,
\[ \Delta t= \frac{\mbox{distance travelled in time interval}}{\mbox{speed}}= \frac{\mbox{\rm arc PQ}}{v} = \frac{r\phi}{v} \]
where the last step uses the formula for the length of the arc of a circle. Combining these results we find
\[ \frac{\Delta v}{\Delta t} = \frac{v^2}{r} \cdot \frac{2 \sin{\phi/2}}{\phi} \]
Finally, as Δt approaches zero so does φ and
sinφ/2φ/2
The magnitude of the acceleration follows as
\[ a = \frac{v^2}{r} \cdot \frac{2 \times \phi/2}{\phi} = \frac{v^2}{r} \]
which completes the argument.