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.Hint
Reveal
Solution
Reveal

Solution
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 (the velocity at P)
and
(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:
![\[ \Delta \vec{v} =\vec{v}_P - \vec{v}_Q \]](mastermathpng-2.png)
The average acceleration over this interval (of duration Δt, say) is then (recall Equation 1.4, written for the case of 1D motion)
![\[ \vec{a}_{\mbox{\rm av}} = \frac{\Delta \vec{v}}{\Delta t} \]](mastermathpng-3.png)
![\[ \vec{a}= \lim_{\Delta t \rightarrow 0} \frac{\Delta \vec{v}}{\Delta t} \]](mastermathpng-4.png)
By inspection we can see that the direction of is that of
, and thus of
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 we use a little trigonometry on the
vector triangle, to give
![\[ \Delta t= \frac{\mbox{distance travelled in time interval}}{\mbox{speed}}= \frac{\mbox{\rm arc PQ}}{v} = \frac{r\phi}{v} \]](mastermathpng-9.png)
![\[ \frac{\Delta v}{\Delta t} = \frac{v^2}{r} \cdot \frac{2 \sin{\phi/2}}{\phi} \]](mastermathpng-10.png)
![\[ a = \frac{v^2}{r} \cdot \frac{2 \times \phi/2}{\phi} = \frac{v^2}{r} \]](mastermathpng-11.png)