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:
- But from the theory of the centripetal acceleration
(Key Point 1.8) we know that
Combining these equations we deduce that
- Finally we can read off the tangential acceleration as