Q2.10 Centripetal and tangential acceleration (S)
The figure shows the acceleration and velocity vectors
of a particle moving in a circle of
radius r=2.5 m, at a particular instant. The magnitude of
the acceleration vector is a=15ms-2, and the angle α=30∘.
At this instant:
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Hint
Reveal
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Hint
You may be more used to determining the centripetal acceleration from the speed; here you just have to go round the loop the other way. But first you need to read off the centripetal (radial) component of the acceleration vector.Solution
Reveal
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Solution
- What is happening? Well the direction of the acceleration vector tells us that there is a component of the acceleration in the same direction as the velocity vector. So one can say that the particle is going round in a circle, and gathering speed at this point.
- We can read off the centripetal acceleration
as the component of the vector
directed towards the centre:
![\[ a_{c} = a \cos(\alpha) = 15\frac{\sqrt{3}}{2} =13 \, ms^{-2} \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]](mastermathpng-1.png)
- But from the theory of the centripetal acceleration
(Key Point 1.8) we know that
Combining these equations we deduce that
![\[ a_c = \frac{v^2}{r} \]](mastermathpng-2.png)
![\[ v= \sqrt{r a_c} = 5.7\, ms^{-1}\hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]](mastermathpng-3.png)
- Finally we can read off the tangential acceleration as
![\[ a_t=a \sin(\alpha) = 7.5 \, ms^{-2}\hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]](mastermathpng-4.png)
