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:
  1. Describe what is happening, in words.
  2. Find the centripetal acceleration.
  3. Find the speed.
  4. Find the tangential acceleration.
Graphic - No title - centandtangaccn
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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

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Solution

  1. 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.
  2. We can read off the centripetal acceleration as the component of the vector $\vec{a}$ directed towards the centre:
    \[ a_{c} = a \cos(\alpha) = 15\frac{\sqrt{3}}{2} =13 \, ms^{-2} \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]
  3. But from the theory of the centripetal acceleration (Key Point 1.8) we know that
    \[ a_c = \frac{v^2}{r} \]
    Combining these equations we deduce that
    \[ v= \sqrt{r a_c} = 5.7\, ms^{-1}\hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]
  4. Finally we can read off the tangential acceleration as
    \[ a_t=a \sin(\alpha) = 7.5 \, ms^{-2}\hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]